Parametric Equation Differentiation

In summary: I mean -2sin³t costidthis is a bit of a pain in the ass but you can do it!Thanks :)In summary, the homework statement is that 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.
  • #1
thomas49th
655
0

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 :)
 
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  • #2
thomas49th said:
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
dy/dt = 2cos²t
dx/dt = -2cosec²t

is that right?
 
  • #4
thomas49th said:
dy/dt = 2cos²t
dx/dt = -2cosec²t

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

dx/dt is right.
 
  • #5
4cos²t ?

Is that right now

Thanks
 
  • #6
No!

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

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

= 2sinxcosx

now where do i go?

Thanks
 
  • #8
Think how to get [tex] \frac{dy}{dx} [/tex] on its own. Basically eliminating the [tex] dt [/tex].
Any ideas?
 
  • #9
thomas49th said:
[tex]4\sin t\cdot\frac{d}{dt}\sin t[/tex]

= 2sinxcosx

now where do i go?

Thanks
Your constant is wrong.
 
  • #10
Now you answer tiny-tims original question:
tiny-tim said:
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
dx/dt = -2cosec²t
dy/dt = 4sin(t)cos(t)now i don't know how to lay it out but i remeber somthing about flipping dy/dx around and everyone going you can't do that in the class. Can you show me?

Thanks
 
  • #12
thomas49th said:
dx/dt = -2cosec²t
dy/dt = 4sin(t)cos(t)


now i don't know how to lay it out but i remeber somthing about flipping dy/dx around and everyone going you can't 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
Try dividing one equation by the other, but which one?
 
  • #14
(dy/dt) / (dx/dt)
= dy/dt * dt/dx
= dy/dx

yes yes. so now where. That was fun!

Thanks
 
  • #15
thomas49th said:
That was fun!

Maths is fun sometimes! :biggrin:
so now where.

So now apply your fun formula to:
thomas49th said:
dx/dt = -2cosec²t
dy/dt = 4sin(t)cos(t)

:smile:
 
  • #16
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
No problem.
 
  • #18
thomas49th said:
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
tiny-tim said:
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 :)
 

What is a parametric equation?

A parametric equation is a mathematical representation of a curve or surface in which the coordinates of a point are expressed in terms of one or more parameters, such as time or distance.

What is differentiation?

Differentiation is a mathematical process that involves finding the rate of change of a function with respect to its input variable. It is also known as the slope or derivative of a function.

How is differentiation applied to parametric equations?

To differentiate a parametric equation, we use the chain rule, where the derivatives of the parametric variables are multiplied by the derivatives of the corresponding functions in the equation.

Why is differentiation important in parametric equations?

Differentiation allows us to find the instantaneous rate of change of a parametric curve at any given point. This is useful in many fields, such as physics, engineering, and economics, where understanding the rate of change of a variable is crucial.

What are some real-world applications of parametric equation differentiation?

Parametric equation differentiation has many practical applications, including motion analysis, calculating velocity and acceleration of objects, optimization problems, and modeling biological processes such as population growth.

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