MHB Neonblast342's question at Yahoo Answers regarding finding a rate of change

[MarkFL]

Here is the question:

Calculus problem! Please help!?

The area of a sector in a circle is given by the formulawhere r is the radius andis the central angle measured in radians. Find the rate of change ofwith respect to r if A remains constant. What is the rate when r = 6?

Here is a link to the question:

Calculus problem! Please help!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello neonblast342,

The formula for the area of the described circular sector is:

$\displaystyle A=\frac{1}{2}r^2\theta$

where $r,\theta>0$

We need to compute $\displaystyle \frac{d\theta}{dr}$.

I would first multiply through by 2:

$\displaystyle 2A=r^2\theta$

Now, differentiate with respect to $r$, and since $\theta$ is a function of $r$, we must use the product rule on the right side. We should recall that $A$ is a constant:

$\displaystyle 0=r^2\frac{d\theta}{dr}+2r\theta$

If we observe that $0<r$, otherwise we have a degenerate sector where $\theta$ has no meaning, then we may divide through by $r$ to obtain:

$\displaystyle 0=r\frac{d\theta}{dr}+2\theta$

Solve for $\displaystyle \frac{d\theta}{dr}$:

$\displaystyle \frac{d\theta}{dr}=-\frac{2\theta}{r}$

The negative sign indicates that $r$ and $\theta$ must move in opposite directions in order for $A$ to remain constant. IN other words, if $r$ increases, the $\theta$ must decrease and vice versa.

We want to have our derivative in terms of $r$ alone, so using the formula for the area $A$ of the sector, we find:

$\displaystyle \theta=\frac{2A}{r^2}$

and so we have:

$\displaystyle \frac{d\theta}{dr}=-\frac{2\frac{2A}{r^2}}{r}=-\frac{4A}{r^3}$

When $r=6$ we find:

$\displaystyle \frac{d\theta}{dr}|_{r=6}=-\frac{4A}{6^3}=-\frac{A}{54}$