Question 1A: Find all critical points of the function...

Math 53: Fall 2020, UC Berkeley Lecture 011 Copyright: J.A. Sethian

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Question 1A: Find all critical points of the function

Set both = 0

Solve for x and y: y= -2x, so x + 2 (-2x) + 1 =0, so -3x=-1 so x=1/3, y =-2/3

Test critical point: fxx= 2, fyy= 2, fxy = 1, so D = fxx fyy – (fxy)2 = (2)(2) -1 = 3

So (1/3, -2/3) is a global and local minimum: Value is (1/3)2 + (1/3)(-2/3) + (-2/3)2 + (-2/3)

Question 1B: Find all critical points of over the square [-1,1]x[-1,1]

We must check both the interior of the square and the boundary

Interior: Set both = 0, So (0,0) is a critical point

Test critical point: fxx= 2, fyy= -2, fxy = 0, so D = fxx fyy – (fxy)2 = (2)(-2) -0 = -4

So (0,0) is a saddle point:

Boundary: global minimum at (0,1), (0,-1) global max at (1,0), (-1,0)



Question 2: Find the cosine of the angle between the planes

Normal to the first plane is u=(1,2,2). Normal to the second plane is v=(1,4,8)

Cosine of the angle between the two is



Question 3: Find the angle made by the two tangents to the curve x=sin t and y = sin t cos t at the point (0,0)

Draw the curve

At intersection point (0,0), we have either t=0 or t = p

At t=0: dy/dx = 1 At t= p: dy/dx = -1

Compute tangent:

So angle is p/2



Question 4: Suppose you are told that there is a twice differentiable function f(x,y,z) with continuous second derivatives, such that

Without using antidifferentiation (that is, without using integration), explain why this cannot be true.

By Clairaut’s theorem, we must have that fxy = fyx

fxy = 3 f yx =2 However,



Question 5a: If and

Find in terms of t

Question 5b: Suppose where f is a differentiable function. What is



Question 6: Let f=sin(xy). Find the unit vector that points in the direction of maximum rate of change of f at the point (1,0).

Direction of maximum change is gradient

At input (1,0), this is (0,1)



Question 7: Let g(z) be a function from R to R Let z=f(x,y) be a function from R2 to R Let C be a level curve of f, parameterized by t, so that C is given by (u(t),v(t)) Let w(t) = g(f(u(t),v(t))) Find the numerical value of

Since C is a level curve of f, then as you traverse (u(t),v(t)), you always get the same output for f(u(t),v(t))

So g(f(u(t),v(t))) always hands out the same number for any value of t

So



Question 8: Find the equation of the plane tangent to x2 + 3y2 + 6z2 = 67 at the point (1,2,3)

Let g(x,y,z) = x2 + 3y2 + 6z2

Then, at the input (1,2,3), the normal to the tangent plane is given by the gradient

Evaluated at (1,2,3), the normal is then (2,12,36)

So the tangent is given by (2,12,36) dot (x-1,y-2,z-3) = 0 which is 2(x-1) + 12 (y-2) + 36(z-3) = 0



Question 9: Who lives in a pineapple under the sea?



Question 10: Consider the function xyz=k, where k is a constant bigger than zero. For any point (x0,y0,z0) on that surface, you can construct a plane tangent to the surface at the point.

Let A be the place that plane hits the x axis. Let B be the place that plane hits the y axis. Let C be the place that plane hits the z axis.

Show that the product of A, B, and C is the same, regardless of which point (x0,y0,z0) you pick on the surface.

At the input (x0,y0,z0), the gradient of g(x,y,z)=xyz is

So the tangent plane passing through (x0,y0,z0) is

A is where this plane hits the x axis, so at A we have y=z=0, A =

B is where this plane hits the y axis, so at B we have x=z=0, B =

C is where this plane hits the z axis, so at C we have x=y=0, C =

So the product ABC =



Okay---a review of Lagrange Multipliers and some problems:

Suppose we want to find the extreme points of a function f(x,y,z) subject to the constraint g(x,y,z)=k. Then at an extreme point we must have that they point in the same direction, so we must have that



More examples: Use Lagrange multipliers to find the maximum, and minimum values of f subject to the constraint

So points are

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Question 1A: Find all critical points of the function...

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