# Find dy/dx as a function of t for the parametric equations

## Homework Statement

find dy/dx as a function of t for the parametric equations
x=cos^7(t)
y=6sin^2(t)

## The Attempt at a Solution

well i'm looking for dy/dx.. so first i found dy
dy = 12sin(t)cos(t)

and dx
dx = -7cos^6(t)sin(t)

dy/dx = 12sin(t)cos(t) / -7cos^6(t)sin(t)
(-12/7cos^5(t))

And then it asks for if the curve is concave up or concave down (besides where it is undefined). When i graph it on wolfram alpha it gives both up and down concaves so it leads me to believe I did this wrong. Help

CompuChip
Homework Helper
dy/dx is the derivative of y with respect to x, it is not technically a fraction. In many cases you can treat it as such, but this is one of those cases where it doesn't really work.

Does the chain rule ring a bell when written as
$$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt}$$
or, more conveniently for you,
$$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \text{ ?}$$

eumyang
Homework Helper
To find the concavity you need the second derivative. I only see the first derivative in your post. Also, when you say that you found "dy" and "dx", you should say instead "dy/dt" and "dx/dt".

EDIT: Regarding my second point, CompuChip beat me to it. :tongue:

dy/dx is the derivative of y with respect to x, it is not technically a fraction. In many cases you can treat it as such, but this is one of those cases where it doesn't really work.

Does the chain rule ring a bell when written as
$$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt}$$
or, more conveniently for you,
$$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \text{ ?}$$

It does not. I just started calculas 3 and we are going over alot of calculus one stuff which i got a D in over 5 years ago. lol ;-(. Calc 2 i stepped up my game and got an A but alot of this stuff is confusing me. I'll go relearn the chain rule. That dt is some kind of new variable that comes out of using the chain rule?

dy/dx = (dy/dt) / (dx/dt). I don't understand how this will get me a result different from what I did the first time. Isn't dy/dt 12sin(t)cos(t)? and dx/dt is 7cos^6(t)sin(t)?

eumyang
Homework Helper
dy/dx = (dy/dt) / (dx/dt). I don't understand how this will get me a result different from what I did the first time. Isn't dy/dt 12sin(t)cos(t)? and dx/dt is 7cos^6(t)sin(t)?
I didn't say that your first derivative was wrong (although you forgot a negative in dx/dt in your more recent post). I'm saying that to test for concavity you need to find the second derivative.

1 person
pbuk
Gold Member
dy/dx = (dy/dt) / (dx/dt). I don't understand how this will get me a result different from what I did the first time.

It won't - in this problem. But if you start writing things like dy = 12sin(t)cos(t)
1. it looks like you don't know what you are doing
2. when problems get harder (like in the second half of this question) you will get caught out

As already stated you need to look at ## \frac{d^2y}{dx^2} ## to get the curve's concavity.

okay.
so dy/dx = -12/(7cos^5(t))
I need to take the derivative again to find the concavity
(d/dt)(dy/dx) = (d/dt)(-12/(7cos^5(t)) = (-60/7)tan(t)sec^5(t). And yet when i graph this curve, it looks like it has sections where it concaves up and some where in concaves down.

alright so i see my above post is wrong.. Now i have
(d/dt)(dy/dx) / (dx/dt) = the second derivative i'm looking for.. after doing all the math i got (-60/49)sec^12(t).. anyone have advice?

I GOT IT thanks everyone :D