My Math Forum Application of differentiation

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 April 18th, 2009, 09:15 PM #1 Newbie   Joined: Apr 2009 Posts: 14 Thanks: 0 1. The area between two varying concentric circle is at all times 9phi in^2. the rate of change of the area of the larger circle is 10phi in^2/sec. How fast is the circumference of the smaller circle changing when it has area 16phi in^2 2. A wall of a building is to be braced by a beam that must rest on the ground and pass over a vertical wall 10ft high that is 8 ft from the building. find the length L of the shortest beam that can be used. Last edited by skipjack; April 16th, 2015 at 07:06 PM.
 April 19th, 2009, 12:41 PM #2 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 Re: Application of differentiation I'll work out the answer to the first problem for you. First, we give variable names to all the quantities under consideration: $A$ and $a$ for the area of the larger and smaller circle respectively, $R$ and $r$ for the radii, $C$ and $c$ for the circumferences. As it turns out, we will not need to consider $R$, but we have the formulas $\begin{array}{rclrcl} A &=& \pi R^2 & C &=& 2\pi R \\ a &=& \pi r^2 & c &=& 2\pi r \\ A\,-\,a &=& 9\phi\,\mbox{in}^2 \\ \frac{dA}{dt} &=& 10\phi\,\frac{\mbox{in^2}}{\mbox{sec}}. \\ \end{array}$ Now, we find the circumference of the smaller circle in terms of the area of the larger: $c\,=\,2\pi r\,=\,2\pi\sqrt{\frac{a}{\pi}}\,=\,2\sqrt{\pi a}\,=\,2\sqrt{\pi(A\,-\,9\phi)}$ and differentiate by $t$: $\frac{dc}{dt}\,=\,\frac{dc}{dA}\frac{dA}{dt}\,=\,2 \cdot\frac{1}{2}(\pi(A\,-\,9\phi))^{-\frac{1}{2}}\cdot \pi \cdot\frac{dA}{dt}\,=\,\sqrt{\frac{\pi}{A\,-\,9\phi}}\cdot \frac{dA}{dt}\,=\,\sqrt{\frac{\pi}{a}}\cdot \frac{dA}{dt}.$ At $\frac{dA}{dt}\,=\,10\phi\,\frac{\mbox{in}^2}{\mbox {sec}}$ and $a\,=\,16\phi\,\mbox{in}^2$, we have $\frac{dc}{dt}\,=\,\sqrt{\frac{\pi}{16\phi}}\cdot 10\phi\,=\,\frac{5}{2}\sqrt{\pi\phi}\,\frac{\mbox{ in}}{\mbox{sec}}.$
April 15th, 2015, 08:11 AM   #3
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 Originally Posted by Tear_Grant 1. The area between two varying concentric circle is at all times 9phi in^2. the rate of change of the area of the larger circle is 10phi in^2/sec. How fast is the circumference of the smaller circle changing when it has area 16phi in^2
I assume you mean pi ($\pi$) , and not phi ($\phi$)

Quote:
 The area between two varying concentric circle is at all times 9phi in^2.
$\displaystyle \pi(R^2-r^2) = 9\pi$

$\displaystyle R^2 - r^2 = 9$

$\displaystyle \frac{d}{dt}\left(R^2 - r^2 = 9\right)$

$\displaystyle 2R\frac{dR}{dt} - 2r\frac{dr}{dt} = 0$

$\displaystyle R\frac{dR}{dt} - r\frac{dr}{dt} = 0$

Quote:
 the rate of change of the area of the larger circle is 10phi in^2/sec.
$\displaystyle A_R = \pi R^2$

$\displaystyle \frac{dA_R}{dt} = 2\pi R \frac{dR}{dt} = 10 \pi \implies R\frac{dR}{dt} = 5$

Quote:
 How fast is the circumference of the smaller circle changing when it has area 16phi in^2
$\displaystyle 16\pi = \pi r^2 \implies r = 4$

$\displaystyle C_r = 2\pi r$

$\displaystyle \frac{dC_r}{dt} = 2\pi \frac{dr}{dt}$

looks like you need to find the value of $\displaystyle \frac{dr}{dt}$ when $\displaystyle r = 4$ ... you have enough information above to determine that value.

Last edited by skipjack; April 16th, 2015 at 07:42 PM.

April 15th, 2015, 10:15 AM   #4
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 2. A wall of a building is to be braced by a beam that must rest on the ground and pass over a vertical wall 10ft high that is 8 ft from the building. find the length L of the shortest beam that can be used.
Note the sketch for the assignment of variables.

Pythagoras ...

$\displaystyle L^2 = (x+8 )^2 + (y+10)^2$

similar triangles ...

$\displaystyle \frac{y}{8} = \frac{10}{x}$

use the similar triangles proportion to solve for y in terms of x (or x in terms of y, your choice) and substitute into the Pythagoras equation to get $L^2$ in terms of a single variable.

let $\displaystyle L^2 = Z$ ...

$\displaystyle Z = (x+8 )^2 + (y+10)^2$

note that minimizing $Z$ will also minimize $L$.

find $\displaystyle \frac{dZ}{dx}$ or $\displaystyle \frac{dZ}{dy}$ and minimize.
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