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

• PsychonautQQ
In summary, the homework statement asks for the derivative of y with respect to x, but in this problem, to find the concavity, the student needs to look at the second derivative.

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

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,
$$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \text{ ?}$$

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:

CompuChip said:
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 a lot 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 a lot 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)?

PsychonautQQ said:
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
PsychonautQQ said:
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

1. What are parametric equations?

Parametric equations are a set of equations that express a set of variables as functions of one or more independent variables, typically denoted as t or θ. These equations are commonly used in math and physics to represent curves or other complex shapes.

2. How do you find dy/dx for parametric equations?

To find dy/dx for parametric equations, you can use the chain rule. First, rewrite the equations in terms of x and y. Then, take the derivative of both equations with respect to t. Finally, use the chain rule to find dy/dx by dividing the derivative of y by the derivative of x.

3. Can you provide an example of finding dy/dx for parametric equations?

Sure! Let's say we have the parametric equations x = 2t and y = t^2 + 1. First, rewrite them in terms of x and y: t = x/2 and y = (x/2)^2 + 1. Taking the derivative with respect to t, we get dx/dt = 1/2 and dy/dt = x/2. Using the chain rule, we get dy/dx = (dy/dt)/(dx/dt) = (x/2) / (1/2) = x.

4. What is the significance of finding dy/dx for parametric equations?

Finding dy/dx for parametric equations allows us to determine the slope of the curve at any given point. This can be useful in analyzing the behavior of the curve and making predictions about its direction and shape.

5. Are there any other methods for finding dy/dx for parametric equations?

Yes, there is another method called the geometric interpretation method. This method involves drawing a tangent line at a specific point on the curve and finding the slope of that line. However, this method can be more time-consuming and less accurate than using the chain rule.