welcome to calculus three video lecture number 16 on directional derivatives and the gradient vector so if you'll recall partial derivatives with respect to X or Y provide the rate of change in the direction of either the x axis or the y axis so for example see we have a flat plate here and the temperature of that flat plate is given by a function of x and y the location where you are on the plate okay then if we looked at the partial derivative with respect to X that would give us the rate of change in temperature in the horizontal direction and the partial with respect to Y gives us the rate of change of the temperature in the vertical direction but what if I wanted to move in some other direction and know what the rate of change in the temperature was going to be why do we have to restrict ourselves to only moving in the horizontal or vertical direction well in that case we're gonna have to figure out how to find a directional derivative so we know that we can describe directions using unit vectors and so suppose I want to know what the derivative or the rate of change of the temperature is on this plate in the direction of some arbitrary unit vector with components a B okay so the process is as follows I'm going to draw a line that contains that vector u so I'm going to extend it this is my line let's call it out and I'm gonna locate another point on the line so here is P that's where I'm interested in figuring out the rate of change of the temperature and I'm gonna pick another point Q on the line well I know this line out has the same direction as our unit vector U and this vector PQ notice here's PQ that's a vector it's a scalar multiple of the unit vector u right they're parallel they have the same direction just PQ maybe a different links so I'll say H times u is how I can express PQ H is a scaler and if I were to look at the magnitude of PQ that would equal the magnitude of H the scalar times U which would just equal the absolute value of H because we know the magnitude of U is 1 alright so if I say any point P has coordinates X naught and why not how could I figure out the coordinates now of a point Q well how would you get to Q well remember I'm saying that PQ is equal to H times U so qs x coordinate would be x naught plus h times a remember the components of you were a and b and then the y component would be y naught plus h times B right because this distance here and this distance here would get us to Q so now say I'm interested in the average rate of change as we move from P to Q average rate of change of the temperature right well I would have Delta T over H and that would give me well delta T would be T of Q minus T of P divided by H and then T of Q or I'm gonna use the fact that we said T was a function of x and y so I could also express that as 1/2 remember the coordinates for Q we found them right here here's Q's coordinates so that would be f of X naught plus h a why not plus HB minus now T of P remember P's coordinates are just excellent and why not so this would be f of X naught Y naught divided by H okay well this is the average rate of change so remember to get the instantaneous rate of change which would be our derivative I need to do one more thing so to get the instantaneous rate of change at P in the direction of U our unit vector we would need to take the limit as H approaches zero of this quantity and now we have the definition for our directional derivative so the directional derivative of F at X naught Y naught in the direction of a unit vector U with components AP is given as follows so here's the notation you write capital D subscript is U the unit vector so that means your directional derivative U tells you the direction of your function f X naught y naught that's the point it is going to equal the limit as H approaches zero and in the numerator we have that same difference from up above f of X naught plus h a y naught plus HB minus f of X naught y naught over H all right now in practice you're not going to use this limit definition to find the directional derivative much like most of the time we don't use the limit definition to find derivatives anyhow so here is a theorem that allows us to compute it in a much more convenient form and it tells us if our function f is differentiable and it has to be differentiable for x and y then f has a directional derivative in the direction of unit vector with components a and B and it can be found as follows so your directional derivative and say we're also writing it at the point X naught Y naught so the directional derivative of F at X naught Y naught would be equal to the partial derivative with respect to X evaluated at X naught y naught times a flux the directional derivative or the derivative partial derivative with respect to Y at X naught Y naught times B and most textbooks have a really nice straightforward proof of this theorem so I recommend reading through it and interestingly notice this is how our understanding of the partial derivative with respect to x and y comes into a better light here say I wanted a unit vector in the direction of the x-axis well easiest choice would be to let u equal 1 0 right so if you want your directional derivative in the direction of the x axis you would have the partial with respect to x times well a is 1 + this needs to go away fernette plus partial with respect to Y times 0 and you just get partial with respect to X which is why after that X partial with respect to X is the directional derivative in the direction of the x-axis and then I'm sure it's no surprise if you want the directional derivative in the direction of the y-axis you would be 0 1 you apply the same definition you have partial with respect to x times 0 plus partial with respect to Y times 1 and you get your partial with respect to Y so basically the idea of the directional derivative is we don't want to restrict ourselves to only finding rates of change in the direction of the x axis or the y axis we want to be able to pick any direction and so all we need to do is define the unit vector so here we go we're going to look at some examples let's find the directional derivative of f of X y equals x over y at 6 negative 2 so this is my point x naught y naught in the direction of V and V has components negative 1 free well first thing we need to address is the fact that I need a unit vector so the magnitude of V is gonna be rad 10 so in order to make it a unit vector I just need to divide by that so 1 divided by rad 10 times negative 1 3 and so we get negative 1 over r ab 10 3 over 10 that's what we're gonna use free do all right now I need my partial derivatives so partial derivative with respect to X remember holding white constant thought you just give me 1 over Y so that means that my partial derivative with respect to X evaluated at the 6 negative 2 would just be negative 1/2 and then partial derivative with respect to Y we'll remember that's x times y to the negative first power right if I'm shooting Y is the variable so now that's gonna be negative x over Y squared so my partial with respect to Y evaluated at 6 negative 2 that's gonna be negative 6 over positive 4 so negative 3 halves alright so these two values I'm gonna hold on to right now and then I'm gonna put it all together find my value of the directional derivative of F at the point 6 negative 2 so I take negative 1/2 partial with respect to X at the point and then I'm gonna multiply now by a remember the X component of our unit vector which is going to be negative 1 over rad 10 plus and then I take negative 3 halves partial with respect to y at the point and then I multiply that by B which is 3 over Rab 10 the y component of our unit vector and then let's see here so I'm gonna have one over to round 10-9 over to rod ten so that's negative eight over to rod 10 which is negative 4 over rad 10 and that's our final answer for this problem not too terrible right okay so let's look at another example here find the directions in which the directional derivative of f of XY equals x squared plus sine x y at 1 0 has the value of 1 so in this case they're setting it up a little differently than before we have to find the directions meaning we need to figure out what u is and they gave me the value of the directional derivative already so basically we know that the directional derivative of F at the point 1 0 in the direction of U is equal to 1 what I need to do is figure out what is U so we'll say you have components a and B that's what we're after all right what I'm gonna start by finding the partial derivatives and then solving for my unknowns from there so a partial derivative with respect to X derivative of x squared is going to be 2x and then when I take the derivative of sine X Y remember we have to use the chain rule so I'm gonna multiply by Y so I have plus y cosine XY and then partial derivative with respect to Y is gonna be derivative of x squared will give me 0 and then I have X cosine XY good now both of these I need to evaluate at the point 1 0 so partial with respect to X at 1 0 is gonna be 2 times 1 plus 0 so that's just 2 and then partial with respect to Y at 1 0 that's just gonna be 1 times 1 so 1 alright well I know that my partial with respect to X at 1 zero times a plus the partial with respect to Y at one zero times B is equal to and they told me the value of the directional derivative there is one so I can set that equal to one which means now I can replace this first quantity with two right first so with respect to X at 1 0 is 2 times a plus partial with respect to y at 1 0 just 1 times B equals 1 ok so that's good however I have two unknowns and only one equation at the moment which is problematic how can I solve for a and B but I'm not using one other piece of information and that's the fact that U is a unit vector so what does that mean that means that its magnitude is 1 therefore a squared plus B squared has to equal 1 right since you was a unit vector so now I could put everything together and just solve by substitution so I have this cute little system of equations here all right let's separate our work and say I solve for B so I know B is gonna equal 1 minus 2 a and I focus it to that into the second equation so here's the equation 1 here's equation - I'll have a squared plus 1 minus 2 a squared equals 1 which gives me a squared plus 1 minus 4a plus 4a squared equals 1 so the ones cancel out then I have 5a squared minus 4a is equal to 0 factor out the a and I have 5a minus 4 and 0 so what does this mean well a is either equal to 0 or a is equal to 4/5 so I have two options for my unit vector unit vector number one if a is zero B has to be one right it's magnitude is one unit vector number two if a is 4/5 then I can solve for B come back here that way you won't have any ambiguity as to the sine of B right so if a is 4/5 so that's gonna be eight fifths so B is one minus 8/5 which is gonna be negative three-fifths and then you can confirm the magnitude should be one which it clearly is okay so there's our solution right there pretty fun huh all right now it can also be useful to represent the desired direction by using a unit vector expressed in terms of an angle theta so find the directional derivative of the following function f of X Y at the point P with coordinates one negative one in the direction of the unit vector whose Direction is given by the angle PI over three so basically you just have to think of what are the values on the unit circle at PI over three to figure out the components of the unit vector U so cosine of PI over 3 is 1/2 and sine of PI over 3 is rad 3 over 2 and so those are the components of U and now they just want the value of the directional derivative so let's go ahead and take our partial derivatives first so partial with respect to X is going to be 2x + 6 y partial with respect to Y is going to be 6x minus 6y and then now I want to evaluate each of these partials at the point P so at 1 negative 1 the partial with respect to X is going to be 2 minus 6 so negative 4 and then partial with respect to Y at 1 negative 1 is gonna be 6 minus negative 6 so 6 plus 6 which is 12 all right now putting it all together I have directional derivative of F in the direction of U one negative one is gonna equal so you take negative 4 times a here's a 1/2 plus 12 times rad 3 over 2 this is gonna give me negative 2 plus 6 rapidly or finally on sir beautiful all right moving on have you felt like you were doing a familiar process here when you were computing this directional derivative did you notice you were doing a dot product the whole time and in fact we have another very useful way to represent the directional derivative using a dot product so think about it you take the partial with respect to X you multiply by a and then you add that to the partial with respect to Y times B so one of the vectors would have to be a vector of partial derivatives right partial with respect to X partial with respect to Y and then you take the dot product of that vector with a B where u has components a and B ok so we're gonna try using this notation instead it's just a little cleaner so find the directional derivative of the following function of XY and Z so in this case I'll have to take three partial derivatives so I'll have vectors in three space at one 62 in the direction of the following vector so first things first we need a unit vector don't we people so let's see what's the magnitude here so 9 plus 16 plus 144 that's right it's 169 so I'm gonna divide you divide V by 13 so now I have you which is gonna be 1 13 times 3 4 12 all right now let's find our vector of partial derivatives so I need a partial with respect to X why and Z in this case so partial derivative with respect to X is going to be negative 2xy partial with respect to Y it's gonna be negative x squared and then partial with respect to Z is 3z squared now if I evaluate this at the point P so what you could do is you could just indicate that you're gonna evaluate this at negative or where's the point 1 6 2 right here's the point P so I'm going to evaluate it at 1 6 2 and I get negative 12 negative 1 and then 3 times 2 squared 3 times 4 12 all right so there's my vector save that and then now I'm ready to find my directional derivative and I didn't distribute that 13 on purpose because watch it's gonna be so nice right now so a directional derivative of F at 1 6 2 in the direction of U is given by dot product of negative 12 negative 1 12 with 1 over 13 times 3 4 12 well that 1 over 13 I can just pull it outside of the dot product right and then negative 12 times 3 so taking the product of those first components that's gonna be negative 36 and then same thing with the second components negative 4 and then I have plus 144 so that's negative 40 plus 144 so that's 104 divided by 13 which is 8 there we go now interestingly the vector of partial derivatives that we saw up above that was used in this dot product has a number of applications and it's such a popular vector it even gets its own name so if F is a function of two variables x and y then we say that the gradient of F is the vector function and it's denoted as follows with that upside-down triangle defined by let's we say the gradient of F equals the vector of the partial of F with respect to X and then the partial of F with respect to Y and you can also extend this for more variables okay now we're gonna study this gradient in more depth for the remainder of this lesson so what we want to do is consider how the value of the directional derivative changes as we change you so now that we know the directional derivative can be written as the gradient of F dotted with you we know that we can rewrite this dot product as the magnitude of the gradient of F times the magnitude of U times cosine of theta right let's say does the angle between the gradient of F and you well I know that the magnitude of U is gonna be 1 because U is a unit vector right so I can rewrite this as just the magnitude of the gradient of F times cosine theta theta is the angle between U and the gradient okay so I want to study how is this value of the directional derivative changing as u changes as the direction of U changes so if the direction of U is changing all that would be changing to represent that would be theta right the angle between U and the gradient well let's think what do we know about cosine of theta well it's bounded cosine of theta is equal to 1 when theta is equal to 0 so that means this quantity here is maxed out when theta equals zero so we achieved the max value of the directional derivative okay other extreme cases cosine of theta is negative one when theta equals pi so in that case we achieve the minimum value of the directional derivative when your unit vector U is in the opposite direction of the gradient so when theta is zero that means they're parallel and in the same direction U and the gradient and when theta is high they're still parallel but they're in opposite directions U and the gradient another way that you would say you have the min value of the directional derivative is it could be referred to this is confusing at time as the max rate of decrease okay and then last scenario that's interesting is cosine of theta equaling zero and that happens when theta is PI over two and in that case the directional derivative is constant and that means the function values aren't changing or they don't change okay very good now let's look at nice exciting follow-up theorem what could be better so suppose F is a differentiable function of two or three variables then the max value of the directional derivative is given by the magnitude of the gradient of f of X and it occurs when u has the same direction as the gradient vector given below okay so we informally came to this conclusion because we knew from previous studies that cosine theta maxes out at 1 when theta is equal to 0 but here we have a theorem that guarantees this and so if you're ever asked to find the maximum value of the directional derivative all you have to do is find the magnitude of the gradient isn't that nice so burn this in your brain max value of directional derivative all you need to do is find magnitude of the gradient don't go back to other calc ways of finding maxima and minima and taking extra derivatives not necessary when you're working with the directional derivative okay so let's apply it now the temperature at some point in space XYZ is given by the following function so you have temperature is determined by your location in three-dimensional space and T is measured in degrees Celsius XY and Z in meters so find the rate of change of the temperature at P in the direction toward the point Q let's call it Q point towards 3 negative 3 3 ok so first thing let's make up vector in that direction from P to Q so the components would be 3 minus 2 which is 1 next component is going to be negative 2 and 1 all right well first thing what do I need to do with this guy that's right we got to make it a unit vector so its magnitude is 6 so my unit vector U it's gonna be one over rad 6 times 1 negative 2 1 very nice ok next I need the gradient so gradient of T remember that's the vector of your partial derivatives I'm gonna take partial with respect to X partial with respect to Y partial with respect to Z so partial derivative with respect to X is just gonna be too wide partial with respect to Y is gonna be 2x minus Z and partial with respect to Z is going to be negative Y and then now I need to evaluate the gradient at the point P because they wanted the rate of change at P so that's the point that I'm evaluating it at so at 2 negative 1/2 so that will give me components negative 2 2 times 2 minus 2 is 2 again but we got to write a nice too don't we that's much better and then negative Y that's gonna be positive 1 all right so now we're ready to put it all together I'm gonna take the dot product now of this vector of my gradient vector at P with you and that is the directional derivative of T at P in the direction of U it's a mouthful isn't it so I'll take negative 2 to 1 dot that with 1 over around 6 times 1 negative 2 1 aren't you glad you didn't distribute the rad six right now okay keep it in the denominator keep it in the denominator and then we'll just do the dot product in the numerator so I'm gonna have negative 2 minus 4 plus 1 so what does that give me negative 5 over rad snakes and we need to include units so notice this is the rate of change of temperature and they told me temperature here is measured in degrees Celsius so this is degrees Celsius per meter very good okay moving on in which direction does the temperature increase the fastest at P well as we looked at before the directional derivative achieves its max value when you your unit vector is parallel in the exact same direction to the gradient vector so it increases the fastest in the direction of the gradient of T at P which was to negative 1/2 which has components negative 2 to 1 second part of the problem what is this rate of increase well thank goodness we have that theorem from up above that tells me that the max rate of increase I don't have to write max just the rate of increase is the magnitude of this vector the magnitude of the gradient at to negative 1/2 so if I take the magnitude I'm gonna have square root of 4 plus 4 plus 1 so that's gonna be 3 degrees Celsius per meter very nice now Part C asks an interesting follow-up at the point P if there a direction in which the rate of change in temperature is negative 5 degrees Celsius per meter it's so where and if not why not well remember since I know the max rate of increase is 3 that also means that the max rate of decrease is going to be negative 3 and that occurs when u is opposite in the opposite direction of the gradient so that means no negative 5 degrees Celsius per meter is possible let's write this all out we know that the most rapid decrease or the minimum value of our directional derivative is in the direction opposite to the gradient so opposite of the gradient of T and it's rate would be the opposite of the magnitude of the gradient in this case that's negative three degrees Celsius per meter so no say it with emphasis there we go negative five degrees Celsius per meter is not possible all right make sure you have those concepts under control because the terminology can sometimes be confusing even though the idea really isn't all right moving on to another interesting example suppose you are climbing a hill whose shape is given by the equation Z which equals one thousand minus 0.01 x squared minus 0.02 y squared where x y and z are measured in meters and the positive x-axis points due east while the positive y axis points due north that's pretty standard people so suppose you are at the point with coordinates 50 80 845 so this is giving you X Y and then this is the your elevation so basically you have a hill and the height or the slope of the hill the shape of it is given by that equation there Z is giving you the heights of the hill all right so the first part asks if you walk due south will you start to ascend or descend and at what rate so I'm going to first call this the function f of X and y to maybe make the notation a little bit easier and basically if I'm gonna figure out to find a spend or descending I want to know what is the rate of change of Z right if I walk due south so in the direction of I have to figure out how to write a unit vector that represents going due south and I want to know if I'm ascending then that rate of change of Z is going to be positive right I'm going uphill and then if I'm descending the rate of change of Z that directional derivative should have a negative value all right so let's draw out a coordinate plane here and Y axis points due north x axis points due east and we're at the point 50 80 and then our elevation is 847 okay so what unit vector could you use to represent walking due south what would its components be that's right 0 negative 1 Scott would take you to South then I want to know what the gradient of s is well the gradient of F partial with respect to X is gonna be negative 0.02 X partial with respect to Y is negative 0.04 Y and now I'm gonna evaluate the gradient at the point well what was the point 50 80 so at 50 80 the value of the gradient is going to be negative 0.02 times 50 that's negative 1 and then negative 0.04 times 80 that's negative 3.2 so now I can find the value of the directional derivative of F at 50 80 it's gonna be negative 1 negative 3.2 dot that with my unit vector u 0 negative 1 and I'm gonna get positive 3.2 now remember that represents the rate of change of Z so that would be meters of elevation per change in the XY plane so you could just say meter of distance right that you moved in that right clean all right so are you ascending or descending did you figure it out well since the value of the directional derivative is positive that means you're going up hill Z is increasing so we are ascending alright Part B if you walk Northwest will you start to ascend or descend and then at what rate so Northwest we need a unit vector that takes us in that direction not bad unless you just get jumbled up which way west is it's over here people where the negative x-axis is okay so if you're walking north west from 50 80 what unit vector could you use to say this direction right here well the components would be negative 1 1 but then make sure it's a unit vector so I would divide by rad 2 so you would have components 1 over rad 2 times negative 1 what and then I already have my gradient at P so this directional derivative is gonna be easy we did all that extra work up above so I'm gonna take the dot product with the same gradient negative 1 negative 3.2 with this new direction u 1 over rad 2 times negative 1 1 so this is gonna give me positive 1 minus 3.2 divided by the square root of 2 which is negative two point two divided by Rab - and remember that's meters of elevation per meter of distance so what would you say about this rate of change are you ascending or descending we'll notice it's negative which means we are descending you're going downhill all right next part in which direction is the slope the largest so when is the rate of change a max o in the direction of the gradient so in the direction of the gradient of F which has components negative 1 negative 3.2 second part of the question what is that rate of a cent well we have the theorem that tells us that the actual max rate can be found by taking the magnitude of the gradient so the magnitude is going to be the square root of 1 squared plus 3 point 2 squared which is about three point three five meters remember of elevation per meter of distance okay at what angle above the horizontal does the path in that direction begin okay so what I want to know is what is the angle theta where that's a terrible triangle come on where this rate of ascent this is Z is equal to three point three five right that's giving me the change in Z so that's three point three five meters and then that's per one meter of distance you could think of this us three point three five change in Z over one metre change in the XY plane so this sides one this side here is three point three five I don't like how I label about one this side down here is one and then they want that angle theta that's right in there okay so tangent of theta is equal to three point three five over one so theta is approximately seventy three point or degrees all right what a nice little example now we're gonna look at how the gradient can be used to simplify finding tangent planes to surfaces remember we talked about that a few lessons ago and so now we want to revisit the topic now that we know something about the gradient okay now assume s is a surface and it satisfies that the funk it's a function of XY and z equal tests some constant K and say we have a point P with coordinates X not Y not zina and that's a point on the surface s and now C is any curve which I can represent parametrically so I'll call it R of T with components x y&z all functions of T and this curve C lies on the surface ok now say I want to differentiate F with respect to T ok so I have my function of XYZ equal to K and now I want to differentiate it with respect to T well remember since F is a function of XY and Z I'm going to have to apply the chain rule so I'll have del F del X times now X is just a function of T so that would be DX DT not a partial plus del F del Y times dy DT plus del F del Z times DZ DT is equal to 0 because K is a constant ok we'll look at this derivative here doesn't it remind you of the dot product just a little bit it sure reminds me of a dot product look let's see if we can come up with the two vectors that were dotted it looks like one of them was del F del X del F del Y del F del Z and what's the other one dx/dt dy/dt and dz/dt this equals zero that's interesting who's this this is our good friend the gradient that's the gradient of F dotted with well remember our my curve had components x y and z so this vector here is just R prime of T and interestingly the gradient of F dotted with our prime of T gives me zero what does that mean they're orthogonal so this tells me that the gradient of F is orthogonal to the tangent right our prime will be tangent to any curve on s that goes through P so basically the gradient of F acts as the normal vector to our tangent how exciting okay so that means the tangent plane to the level surface F of XYZ equal in K at some point P is given by using and equal to the gradient of F evaluated at X naught Y naught Z naught as the normal vector and then from there you're gonna make sure you write your final equation in standard linear form ax plus B y plus cz equaling D ok all right so now we can find equations of tangent planes to surfaces not necessarily when Z is defined as a function of x and y so here let's look at an example find the tangent plane to the following surface at negative 1 1 2 so I'm after that normal vector which is going to be the gradient and since this is equal to a constant I'm just gonna go ahead take my partial derivative so partial with respect to X is going to be 2x partial with respect to Y is gonna be negative 4y and then partial with respect to Z is 2z and then now I need to evaluate that at the point P negative 1 1 negative 2 so first component is gonna be negative 2 negative 4 and negative 4 well I could use that or I can also use for n positive 1 to 2 which is just a little nicer to work with so let's go ahead and do that so I'll have 1 times X plus 1 remember I'm using this point here at the point on the plane plus 2 times y minus 1 plus 2 times Z plus 2 is equal to 0 and then we got X plus 2 y plus 2z is equal to negative 3 all right pretty straightforward now another term here is the normal line we say the normal line to a surface at some point P is the line that goes through P and it's perpendicular to the tangent plane so therefore if this normal line is perpendicular to the tangent plane its direction vector is going to be in the same direction as that normal vector the gradient vector so the normal line could be found by taking the gradient of F at P as the direction vector and then use P of the point to write the parametric equations for this normal line so it's looking at an example pretty straightforward same surfaces before now I want to write equation of the normal line well remember we have the gradient of F at that point from earlier and we found it to be negative 2 and they get it for and they get it for you could use that or it's a lot nicer just to work with 1 to 2 and then I can go ahead write the parametric equations for the normal line so X is gonna equal remember I take that first component and I'll have multiply that by T so I'll have t minus 1 y is gonna equal to t plus 1 remember I'm using the point there and Z is equal to 2t minus 2 so here are the parametric equations for that normal line alright just to tie this all in to when we talked about tangent planes before we said that if Z was a function of x and y then the tangent plane was found by looking at the normal vector with components partial with respect to X partial with respect to Y and negative 1 but now we see that this was just a special case of our new approach because if Z is a function of x and y I can redefine this as a function of x and y in Z I'll call it capital f of X Y Z and remember it needs to equal a constant so I could write f of X Y minus C is equal to 0 and now if I take the gradient of F I would have the partial with respect to X partial with respect to Y and negative 1 which is my normal vector so we were just dealing with a special case before so look at this gradient it has so many uses let's just summarize what we've considered so far the gradient is useful for well it gives us the normal vector right for a tangent plane what else it gives us the direction vector for the normal light it also gives us the direction for the greatest rate of increase and the opposite of the gradient gives the direction for the greatest rate of decrease oh there's more the magnitude of the gradient gives that fastest rate of increase not baddest fastest and the opposite of the magnitude of the gradient gives us the fastest rate of decrease change this to decrease okay also there's more the gradient is orthogonal to level curves remember we showed that the gradient dotted with our prime of T was equal to zero and the gradient is used to find directional derivatives the summary is basically in Reverse this was the first thing we considered remember directional derivatives from the very beginning of the video okay so last example way to go so find the points on the surface where the tangent plane is parallel to this plane here okay well what's the normal vector for this plane the normal vector has components 3 negative 1 3 all right all right so I want to know where is the gradient of this surface the gradient of F equal to K times n where K is a scalar okay well the gradient of F we can find that easily that's 2x4 y6z how about instead we use divide that whole thing like two and just use X 2 y 3z right why not so I want to know when is that equal to K times the normal vector for this other plane because then those two planes would be parallel all right well let's see we have a little system going on here I can set X equal to 3 K let's do that so X is equal to 3 K 2 y would equal negative K and also 3 Z with equal 3 K and now I want to make sure this point is on the surface so I'm gonna say well I'll keep X is equal to 3 K Y is equal to negative K over 2 and Z is equal to K so I'm gonna substitute these in for x y&z into the surface I'll find K first and then I'll find the points where the tangent plane is parallel to the other plane ok so we're plugging all of this in 2 x squared plus 2 y squared plus 3 Z squared is 1 so I'm gonna have X was 3 K right so 3 K squared plus 2 times negative K over 2 squared plus 3 times Z is just K squared is one so what is that 9 K squared plus 1/2 K square plus another 9 K squared so this is gonna give me 25 over now excuse me I have 9 KS squared that one's gonna be 1/2 and then just a three right so 25 over 2 K squared equals 1 which means that K squared equals 2 over 25 so K equals plus or minus grab 2 over 5 all right well we have a couple possibilities then so let's list everything out nicely to begin if K is equal to positive rad 2 over 5 then X would be 3 times that so 3 rub 2 over 5 Y is negative 2 negative rad 2 over 10 and then Z would be equal to K so rad 2 over 5 now if K is equal to negative rad 2 over 5 then X would be negative 3 rub 2 over 5 y would be positive rad 2 over 10 and Z would be negative rad 2 over 5 so that's one option of how to write it out you can't just put plus or minus across but I'll show you another way that you could write it if you don't want to write out the two points separately here's your other option it's pretty slick or you could write out your answer ass-blaster minus 3 red 2 over 5 now notice whatever sign the x coordinate had the y coordinate have the opposite sign right see how these are opposites so to indicate that they're always opposite then for the next coordinate output minus plus rad 2 over 10 and then the Z coordinate matches the sign of the x coordinate so that would be plus or minus grab two over five okay but you can't put plus or minus for all of them I love this the less I write the happier I am you know what they mean looks good so that concludes the lesson it was a beast of one but so interesting and so useful