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# Computing a tangent plane

## Video transcript

hey guys so in the last video I was talking about how you can define a function whose graph is a plane and moreover a plane that passes through a specified point and whose orientation you can somehow specify and we ended up seeing how specifying that orientation comes down to certain partial derivative information and first let me just kind of repeat what the conclusion was but I'll put it in more abstract terms since I did it in a very specific example last time so basically if you want some kind of function which gives you a plane that passes through a certain point well first let's say what that point is right let's say the point was X naught Y naught and Z naught so these are just constant values and this is my way of abstractly describing a single point in space using X naught to represent a constant x value Y naught to represent constant Y value that kind of thing so what it is is it's going to be some sort of other constant a multiplied by X minus X naught so this this white X here is the is the variable and then X naught is just a constant let me let me go ahead and go ahead and pick that parenthesis there okay so then we add to that B multiplied by and then B is just some other constant just like a is some other constant multiplied by Y minus y naught and then all of that you add Z naught and now if you just presented this as it is is kind of a lot right there's five different constants going on but really what this is saying is you want something where the partial derivative with respect to X is just some constant and you want to be able to specify what that constant is and similarly the partial derivative with respect to Y is another constant and you just want to ensure that this passes through this point X naught Y naught Z naught and if you imagine plugging in X the variable equals X naught the constant this part goes to 0 similarly plugging in Y naught the constant makes this part go to 0 so this is a way of specifying that when you evaluate the function at X naught Y naught it equals Z naught and that's what makes sure that the graph passes through that point so with that said let's start thinking about how you can find the tangent plane to a graph and first of all let's think about what that what that point is how you specify such a point instead of specifying any three numbers in space because you have to make sure the point is somewhere on the graph you instead only specify two you're basically going to say what's the x-coordinate and it's this case let's say the x-coordinate was like 1 I'll be like 1 and then the y-coordinate which looks about like negative 2 and to make it easier I'm just going to say let's say it is negative 2 then the Z coordinate is specified because this is a graph the Z coordinate is forced to be whatever the output of the function is at 1 negative 2 so this is going to be whatever the output of our function is at 1 negative 2 and F here F is going to be whatever function gives us this graph so maybe I should maybe I should write down the actual function that I'm using for this graph in this case F which is a function of x and y is equal to 3 minus 1/3 of x squared minus y squared ok so this is the function that we're using and you evaluate it at that point and this will give you your your point in three-dimensional space that our linear function that our tangent plane has to pass through so we can start writing out our function right we can say ok so our linear function as a function of x and y it's got to make sure it goes through that 1 and that negative 2 so this is going to be some constant a that we'll fill in in a moment multiplied by X minus that one plus and then be also a constant with it we'll specify in a moment times y minus that negative 2 so it's minus and negative 2 and then the hole the the thing that we add to it is f of 1 negative 2 and let's just go ahead and evaluate that let's say we plug in 1 and negative 2 so if we go up here and we plug in so that'll be 3 minus x equals 1 well 3 minus 1/3 of if x equals 1 1/3 of 1 squared so that's 1/3 1 squared minus and then Y is negative 2 so that will be minus negative 2 squared so that's 3 minus 1/3 minus 4 so the whole thing is equal to let's see 3 minus 4 is negative 1 minus another third is negative 4/3 ok so that's what we add to this entire thing we add negative 4/3 and maybe I should just kind of may clear the separation here so this is our function but we don't know what a and B are those are things that we need to plug in now the whole idea of a tangent plane is that the partial derivative with respect to X should match that of the original function so if we go over to the graph here and start thinking about partial derivative information if we want the partial derivative with respect to X and you imagine moving purely in the X direction here this intersects the graph along some kind of curve and what the partial derivative with respect to X at this point tells you is the slope of the tangent line I'm kind of bad at drawing there this is the slope of the tangent line in that direction of that point so that's what the partial derivative with respect to X is telling you and what you want when you look at the tangent of the tangent plane is that the tangent plane also has that same slope you know you kind of if I line things up here you'd want it also to have that same slope so you can specify over here and say a we want a to be equal to the partial derivative of the function with respect to X evaluated at this one negative two evaluated at that special point one negative two and then similarly B for pretty much the same reasons and I'll draw it out here so let's kind of ah ah let's erase this line so instead of intersecting it with that slice let's see what movement in the Y Direction looks like so in this case it looks like a very steep slope right because in this case the tangent line in that direction is a pretty steep slope and now when we bring in when we bring in the tangent plane it should intersect with that constant constant x-value plane along that same slope made it kind of messy there but you can you can see the the line formed by intersecting these two planes should be that desired tangent and what that corresponds to in formulas is that this B which represents the partial derivative of L L is the tangent plane function that should be the same as if we take the partial derivative of f with respect to y at that point at this point 1 negative 2 and this is stuff that we can compute and that we can figure out so let's just kind of start plugging that in first let me just copy this function because we're going to need it copy and now let's go on down here I'm just going to let's paste it down here kind of in the bottom because that's what we'll need so let's compute the partial derivative of F with respect to X so we look down here the only place where X shows up is in this negative one-third of x squared context so the the partial derivative of F with respect to X is going to be just the derivative of this little guy which is negative we bring down the two negative two thirds of X so when we go ahead and plug in you know x equals one to see what what it looks like when we evaluate at this point that's just going to be equal to negative two thirds so that tells us that a is going to have to be negative two thirds now for similar reasons let's let's go ahead and compute the partial derivative with respect to Y we look down here well the only place that Y shows up in the entire expression is this negative Y squared so the partial derivative of F with respect to Y is equal to just negative 2y negative 2y and now when we plug in y equals negative two what we get is negative two multiplied by negative 2 that's didn't have to be the case that those were the same and that whole thing equals 4 so the partial derivative of F with respect to Y evaluated at this point one negative two is equal to four so if we were to plug this information back up into a formula we would replace a with negative 2/3 we would say negative 2/3 and we would replace B with four replace B with four and that would give us the full formula we wrote the full formula for the tangent plane and this could be kind of a lot to look at it first because we have to specify the input point you know one negative two and then we had to figure out where the function evaluates at that point and then we had to figure out both of the partial derivatives with respect to X and with respect to Y but all in all there's not actually a lot to remember for how you go about computing this looking at the graph actually makes things seem a lot more reasonable because each of those terms has an actual meaning if we look at the 1 and negative 2 that's just telling us the input the kind of x and y coordinates of the input and of course we have to evaluate that because that tells us the Z coordinate that'll put us on the graph corresponding to that point and then to give us a tangent plane you just need to specify the 2 bits of partial differential information and that'll tell you kind of how this graph needs to be oriented and once you start thinking things in that way you know geometrically even though there's a lot going on here there's five different numbers you have to put in each one of them feels like yeah yeah of course you need that number otherwise you couldn't specify a tangent plane this kind of a lot of information required to put it on the appropriate spot so with that I will see you next video