Let w = 5xy + 2yz - 2xz, x = st, y = e^st, z =...

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Transcript

00:01 In this question, we are asked to find the point on the surface.

00:07 So we are provided with the surface x squared subtracted by y is that is equals to one that is closest to the origin.

00:21 So first let us reconstruct this equation as x squared is equals to one plus y is that.

00:31 So first let's recall the formula of distance which is equals to x squared plus y squared plus z squared to the power of one over two as the given point is origin we don't have any x when y when and is it values.

00:53 So since x y and z lies on the surface this distance can be written as distance squared is equals to.

01:09 So we have x squared as one plus y is it so one plus y is it plus y squared plus z squared.

01:18 So this is our distance.

01:20 So i'm going to write this in the form of a four.

01:23 We have only y and z function.

01:25 So let's write this as y squared plus z squared plus y is that less one.

01:32 So next let's find the derivative with respect to y.

01:36 So we'll be having f y is equals to two y less is it and f is that is equals to two is it less y.

01:49 So to find the critical points we have to equate this partial derivatives with zero.

01:57 So to find the critical points i'm going to equate it with zero so we'll be having two y plus is that is equals to zero and two is it plus y is equals to zero...

Question

Let w = 5xy + 2yz - 2xz, x = st, y = e^{st}, z = t^2 Compute \frac{\partial w}{\partial s}(-3, -1) = \frac{\partial w}{\partial t}(-3, -1) =

Question

Let w = 5xy + 2yz - 2xz, x = st, y = e^{st}, z = t^2 Compute \frac{\partial w}{\partial s}(-3, -1) = \frac{\partial w}{\partial t}(-3, -1) =

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