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Parametric Equation Differentiation

  • Thread starter thomas49th
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  • #1
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Homework Statement


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


The Attempt at a Solution



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

Thanks :)
 

Answers and Replies

  • #2
tiny-tim
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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
Hi thomas49th! :smile:

Hint: what is dy/dx in terms of dy/dt and dx/dt? :smile:
 
  • #3
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dy/dt = 2cos²t
dx/dt = -2cosec²t

is that right?
 
  • #4
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dy/dt = 2cos²t
dx/dt = -2cosec²t

is that right?
dy/dy is incorrect don't forget chain rule!!!

dx/dt is right.
 
  • #5
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4cos²t ???

Is that right now

Thanks
 
  • #6
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No!!!

[tex]y=2\sin^2 t[/tex]

[tex]\frac{dy}{dt}=4\sin t\cdot\frac{d}{dt}\sin t[/tex]
 
  • #7
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[tex]4\sin t\cdot\frac{d}{dt}\sin t[/tex]

= 2sinxcosx

now where do i go?

Thanks
 
  • #8
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Think how to get [tex] \frac{dy}{dx} [/tex] on its own. Basically eliminating the [tex] dt [/tex].
Any ideas?
 
  • #9
1,752
1
[tex]4\sin t\cdot\frac{d}{dt}\sin t[/tex]

= 2sinxcosx

now where do i go?

Thanks
Your constant is wrong.
 
  • #10
HallsofIvy
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Now you answer tiny-tims original question:
Hi thomas49th! :smile:

Hint: what is dy/dx in terms of dy/dt and dx/dt? :smile:
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
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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
tiny-tim
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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
Hi thomas49th! :smile:

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? :smile:
 
  • #13
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Try dividing one equation by the other, but which one?
 
  • #14
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(dy/dt) / (dx/dt)
= dy/dt * dt/dx
= dy/dx

yes yes. so now where. That was fun!

Thanks
 
  • #15
tiny-tim
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That was fun!
Maths is fun sometimes! :biggrin:
so now where.
So now apply your fun formula to:
dx/dt = -2cosec²t
dy/dt = 4sin(t)cos(t)
:smile:
 
  • #16
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ahhh i see

[tex]\frac{dy}{dx} = \frac{4\sin t \cos t}{-2cosec^{2}t}[/tex]

which can be simplified to

[tex]\frac{dy}{dx} = \frac{-2\cos t}{\sin t}[/tex]

[tex]= -2 \cot t[/tex]

Is that right?
 
  • #17
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No problem.
 
  • #18
tiny-tim
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ahhh i see

[tex]\frac{dy}{dx} = \frac{4\sin t \cos t}{-2cosec^{2}t}[/tex]

which can be simplified to

[tex]\frac{dy}{dx} = \frac{-2\cos t}{\sin t}[/tex]

[tex]= -2 \cot t[/tex]

Is that right?
erm … 1/cosec²t = sin²t. :rolleyes:

So dy/dx = … ? :smile:
 
  • #19
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erm … 1/cosec²t = sin²t. :rolleyes:

So dy/dx = … ? :smile:
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 :)
 

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