Derivative of polar equation

rinspd

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Sep 19, 2011
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given the polar equation for position function of a object:
x= 2cos(3t)+3sin(t)+5
y= 10sin(2t)+sin(3t)+11

a/ Find the first and second derivative?
b/State the interval of increasing and decreasing.
c/ When is the moving left, right, up, down.
d/ what is the total distance travel 0<t<2pi.
 
given the polar equation for position function of a object:
x= 2cos(3t)+3sin(t)+5
y= 10sin(2t)+sin(3t)+11

a/ Find the first and second derivative?
b/State the interval of increasing and decreasing.
c/ When is the moving left, right, up, down.
d/ what is the total distance travel 0<t<2pi.

What have you tried, where exactly are you stuck?
 
I have trouble finding the second derivative and also the distance travel on 0<t<2pi, my calculator cannot compute the distance
 
Last edited:
Hello, rinspd!

I'll give you the game plans for this problem.
You can work out the details . . .


Given the parametric equations for position function of a object:

. .
\(\displaystyle \begin{array}{ccc}x &=& 2\cos(3t)+3\sin(t)+5 \\ y &=& 10\sin(2t)+\sin(3t)+11\end{array}\)

(a) Find the first and second derivative.

\(\displaystyle \text{Formulas: }\:\begin{Bmatrix} \dfrac{dy}{dx} &=& \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ \\ \dfrac{d^2y}{dx^2} &=& \dfrac{\frac{d}{dt}\!\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \end{Bmatrix}\)

Be sure to follow that second formula very carefully.



(b) State the interval of increasing and decreasing.

The function is increasing (the point is moving up) when \(\displaystyle \dfrac{dy}{dt} \,>\,0\)

The function is decreasing (the point is moving down) when \(\displaystyle \dfrac{dy}{dt}\,<\,0\)




(c) When is the point moving left, right, up, down?

The point is moving left when \(\displaystyle \dfrac{dx}{dt}\,<\,0\)

The point is moving right when \(\displaystyle \dfrac{dx}{dt}\,>\,0\)



(d) Find the total distance travelled for .\(\displaystyle 0<t<2\pi\)

We want the arc length: . \(\displaystyle \displaystyle L \;=\; \int^{2\pi}_0 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt\)
 
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