# Parametric Equation Differentiation

1. Jun 8, 2008

### thomas49th

1. The problem statement, all variables and given/known data
A curve has parametric equaions:

x = 2cot t, y = 2sin²t, 0 < t <= pi/2

Find an expression for dy/dx in terms of the parameter t

3. The attempt at a solution

Not sure where to go. Do i need to make a Cartesian equation first?

Thanks :)

2. Jun 8, 2008

### tiny-tim

Hi thomas49th!

Hint: what is dy/dx in terms of dy/dt and dx/dt?

3. Jun 8, 2008

### thomas49th

dy/dt = 2cos²t
dx/dt = -2cosec²t

is that right?

4. Jun 8, 2008

### rocomath

dy/dy is incorrect don't forget chain rule!!!

dx/dt is right.

5. Jun 8, 2008

### thomas49th

4cos²t ???

Is that right now

Thanks

6. Jun 8, 2008

### rocomath

No!!!

$$y=2\sin^2 t$$

$$\frac{dy}{dt}=4\sin t\cdot\frac{d}{dt}\sin t$$

7. Jun 8, 2008

### thomas49th

$$4\sin t\cdot\frac{d}{dt}\sin t$$

= 2sinxcosx

now where do i go?

Thanks

8. Jun 8, 2008

### Ed Aboud

Think how to get $$\frac{dy}{dx}$$ on its own. Basically eliminating the $$dt$$.
Any ideas?

9. Jun 8, 2008

### rocomath

10. Jun 8, 2008

### HallsofIvy

Staff Emeritus
Now you answer tiny-tims original question:
Remember that, even though "dy/dx" is not a fraction, it is defined as the limit of a fraction, and you can always treat it as if it were a fraction.

11. Jun 8, 2008

### thomas49th

dx/dt = -2cosec²t
dy/dt = 4sin(t)cos(t)

now i dont know how to lay it out but i remeber somthing about flipping dy/dx around and everyone going you cant do that in the class. Can you show me?

Thanks

12. Jun 8, 2008

### tiny-tim

Hi thomas49th!

Draw a diagram …

x = 2cot t, y = 2sin²t, is a curve.

Draw any old curve, and pretend it's the right one.

Now makes some marks on the curve to indicate equal amounts of t.

Draw a right-angled triangle connecting two of those marks … one side along the curve, and the other two parallel to the x and y axes.

Then dy/dx is the gradient (slope) of the long side, isn't it?

And dy/dt is the right-hand side divided by the change in t.

And dx/dt is the bottom side divided by the change in t.

So (dy/dt)/(dx/dt) is the right side divided by the bottom side … which is the tangent of the slope of the long side … which is dy/dx … isn't it?

13. Jun 8, 2008

### Ed Aboud

Try dividing one equation by the other, but which one?

14. Jun 8, 2008

### thomas49th

(dy/dt) / (dx/dt)
= dy/dt * dt/dx
= dy/dx

yes yes. so now where. That was fun!

Thanks

15. Jun 8, 2008

### tiny-tim

Maths is fun sometimes!
So now apply your fun formula to:

16. Jun 8, 2008

### thomas49th

ahhh i see

$$\frac{dy}{dx} = \frac{4\sin t \cos t}{-2cosec^{2}t}$$

which can be simplified to

$$\frac{dy}{dx} = \frac{-2\cos t}{\sin t}$$

$$= -2 \cot t$$

Is that right?

17. Jun 8, 2008

### Ed Aboud

No problem.

18. Jun 8, 2008

### tiny-tim

erm … 1/cosec²t = sin²t.

So dy/dx = … ?

19. Jun 8, 2008

### thomas49th

o yea! I mean -2sin³t cost

easy

Ok. Now for the next bit i need to find the tangent where t = pi/4

For this should i get all trig terms the same using trig identities i know or sub t= pi/4 in straight away.

Thanks :)