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Differentiating Parametric Functions

  • Thread starter danago
  • Start date
  • #1
danago
Gold Member
1,122
4

Given:
[tex]
\begin{array}{l}
x = 2\cos t \\
y = 2\sin t \\
\end{array}
[/tex]

find [tex]
\frac{{dy}}{{dx}}
[/tex]



I started by finding dy/dt and dx/dt

[tex]
\begin{array}{l}
\frac{{dy}}{{dt}} = 2\cos t \\
\frac{{dx}}{{dt}} = - 2\sin t \\
\end{array}
[/tex]

Now, dy/dx = (dy/dt) / (dx/dt)

so:

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

[tex]
\begin{array}{l}
x = 2\cos t \\
\therefore t = \cos ^{ - 1} \frac{x}{2} \\
\end{array}
[/tex]

Substituting that into my dy/dx which is currently in terms of t:

[tex]
\frac{{dy}}{{dx}} = - \cot t = - \cot \cos ^{ - 1} \frac{x}{2}
[/tex]

Now, my book gives a completely different answer, which isnt even in terms of any trigonometric function. Where have i gone wrong?

Thanks,
Dan.
 

Answers and Replies

  • #2
2,063
2
Do you know that sin2t + cos2t = 1? ;)
 
  • #3
danago
Gold Member
1,122
4
Do you know that sin2t + cos2t = 1? ;)
Yep i do. Not quite seeing where it comes into it though :cry:

EDIT: ahhh now ive got it. Square my x and y functions, and then add them, which eliminates the parameter :) Thanks for that.
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
You could also continue on your original course. cot(cos^(-1)(x/2)) can be written without use of trig functions. Draw a right triangle containing an angle cos^(-1)(x/2).
 

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