Question

When is the slope of r = 3cos(theta) horizontal?

Answer 1

Your first task is to differentiate the function, as that quantity will give you the slope. Then, once you let it equal zero, you will be able to find the value of theta that gives you the horizontal slope. How do you differentiate cos(theta)?

Answer 2

I did. It gave me (pi/4, 1.5sqrt(2)) as one of the points, whose slope is horizontal according to the graph, yet it's marked as wrong.

Answer 3

slope = -cot(2theta)
slope is horizontal when cos(2theta) = 0

Answer 4

That doesn't seem quite right to me. The derivative of cos(theta) is -sin(theta), not -cos(2theta). Also, don't forget the constant 3 that stays outside the function (but ultimately doesn't matter when you're finding that particular value of theta).

Answer 5

This is in polar coordinates.

Answer 6

It shouldn't matter; the derivative is invariant under coordinate transformations as long as you're differentiating with respect to the correct variable.

Answer 7

You can view it as just another optimization problem, replacing r with y and theta with x. It's all the same; differentiation doesn't behave differently.

Answer 8

The graph is a circle centered at (1.5,0) with a radius of 1.5 when expressed with respect to x. Logically speaking, the top and the bottom both have a slope of 0.
They are (1.5, 1.5) and (1.5, -1.5)

Getting the angle of these points, I get pi/4 and 3pi/4, and the radius is 3sqrt(2)/2.
So the points should be (pi/4, 3sqrt(2)/2) and (3pi/4, 3sqrt(2)/2).

Answer 9

Oh wait there's the problem, it's (r, theta) not (theta, r)

Answer 10

I see what you mean now, silly me. Glad you figured it out.