Question

find dy/dx and d^(2)y/dx^(2) if possible, and find the slope and concavity (if possible) at the point corresponding to theta= pi/4

Answer 1

x = -5cos(theta) and y = -5sin(theta)

Answer 2

how do i find d^2y/dx^2 and the slope and concavity

Answer 3

Did you find dy/dx? That certainly would help to find the slope.

Answer 4

yeah i got -5cos(theta)/5sin(theta)

Answer 5

\(x(\theta) = -5\cos(\theta)\)

\(y(\theta) = -5\sin(\theta)\)

\(\dfrac{d}{d\theta}x(\theta) = \dfrac{d}{d\theta}\left(-5\cos(\theta)\right) = 5\sin(\theta)\)

\(\dfrac{d}{d\theta}y(\theta) = \dfrac{d}{d\theta}\left(-5\sin(\theta)\right) = -5\cos(\theta)\)

\(\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \dfrac{-5\cos(\theta)}{5\sin(\theta)} = -\cot(\theta)\)

Now, how about that second derivative? It is NOT just y"/x". What is it?

Answer 6

i dont know @tkhunny

Answer 7

sorry i left yesterday my computer kept lagging out

Answer 8

Seriously? Why are you in this class? Do you have a book? Have you been skipping class?

Try \(\dfrac{\dfrac{d}{d\theta}(-5\cos(\theta))}{5\sin(\theta)}\)

Answer 9

lol okay

Answer 10

5sin(theta)/5sin(theta)

Answer 11

Double Seriously? How about 1?

Answer 12

second derivative is 5sin(theta)/5cos(theta)

Answer 13

1?

Answer 14

please dont be mean i am trying to remember this stuff. its been a long time since i been in school

Answer 15

\(\dfrac{5\sin(\theta)}{5\sin(\theta)} = 1\) except \(\theta = k\pi\) for k \(\in \mathbb{Z}\)

Answer 16

I'm never mean.

Answer 17

so the cosine doesnt change

Answer 18

i mean the sine doesnt change to cosine