Hi, everyone. Welcome back to integralcalc.com. Today we’re going to use the gradient vector
to find the equation of the tangent plane to this function here. We’ve been given
the function x^4 + xy + y^2 = 19. The first thing that we want to do is get all parts
of that function onto one side of the equation so we’re going to convert this to f(x,y)
= x^4 + xy + y^2 – 19. So we just subtracted 19 from both sides to get everything onto
one side of the equation. So we’ve got this equation f(x,y) and we need to find the equation
of the tangent plane at the point P which is given to us as (2,-3). So in order to do
that, the first thing we’re going to do is find the gradient vector of this function
f(x,y), then we’ll evaluate the gradient vector aat the point P (2,-3) and that’ll
give us the vector that is normal to the curve at the point P. Then we’re going to use
that vector as well as the point P to find the equation of the tangent plane. So like
I said, the first thing we’re going to do is find the gradient vector and ther way that
we do that is by taking the partial derivatives of f with respect to both x and y.
So let’s go ahead and take our partial derivatives first. The partial derivative of f with respect
to x will be 4x^3 + y + 0 – 0. The reason that the partial derivative of xy with respect
to x is just y is because if we took the derivative of x alone, the derivative of x with respect
to x is just 1. So we’d end up with 1(y) which of course is just y and therefore our
partial derivative is y. Since there’s no x variable involved in the terms y^2 or -19,
the derivative of both of those is just zero. The partial derivative of f with respect to
y term by term will be zero. There’s no y variable involved in the term x^4 so the
derivative of that term is zero. The derivative of xy with respect to y will just be x. The
derivative of y^2 will be 2y and the derivative of -19 will be zero. So we end up with x +
2y. Now the gradient vector is denoted by this
upside down triangle f(x,y). And it’s just the two partial derivatives. So first the
partial derivative with respect to x which is 4x^3 + y, partial derivative with respect
to y so x + 2y. So that’s our gradient vector but we want to evaluate our gradient vector
at the point that we’re interested which is (2,-3). So instead if f(x,y), we say f(2,-3)
and we plug in 2 for x and -3 for y. When we do that, we get 4(2)^3 + (-3), (2) + 2(-3).
And when we simplify that, we get 32 – 3 = 29, 2 + -6 = 4. So this here, (29,-4) is
the vector that’s normal to the curve at the point P. What we really care about is
that this is basically represented here as a,b. We now have all the information we need
to find the equation of the tangent plane. The formula for that is going to be a(x – x
sub 0) + b(y – y sub 0). And if you had a third variable as you often will when
you’re dealing with gradient vectors, if you had a,b,c here, you could say plus c(z
– z sub 0). But in this case, we only have a and b so that’s what we’re worried about.
And then we set that equal to zero. This is going to give us the equation of our tangent
plane. So obviously, we’ll take a and b. And then
x sub 0 and y sub 0 are going to be coming from our point P as 2 and -3, respectively.
So we already know that a is 29, so we’ll say 29 and we leave this x and y in this equation
as they are so x minus our x coordinate here of our point P which is 2 plus b. b is a -4
so we’re actually going to get minus 4(y – y sub 0). Since we have a -3 for y sub
0 and we have this negative sign here, we’ve got a double negative so we’re going to
end up with y + 3 and then that’s all going to be equal to zero. We could leave this just
like it is but it’s probably cleaner to simplify it. So we’re going to multiply
these out. We’re going to get 29x – 58 – 4y – 12 = 0. And when we reorder these
terms, we do the x term first, we’ll get 29x – 4y – 58 -12 = 0 which would give
us 29x – 4y – 70 = 0 but we like to move that constant to the right side so we’ll
add 70 to both sides and end up with a positive 70 over here on the right. And this is our
final answer. This is the equation of the tangent plane at the point (2,-3): 29x – 4y
= 70. So that’s it. That’s our final answer. I hope this video helped you guys and I will
see you in the next one. Bye!