April 18th, 2009, 08:15 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 06:06 PM. 
April 19th, 2009, 11:41 AM  #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: and for the area of the larger and smaller circle respectively, and for the radii, and for the circumferences. As it turns out, we will not need to consider , but we have the formulas Now, we find the circumference of the smaller circle in terms of the area of the larger: and differentiate by : At and , we have 
April 15th, 2015, 07:11 AM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 2,981 Thanks: 1574  Quote:
Quote:
$\displaystyle R^2  r^2 = 9$ $\displaystyle \frac{d}{dt}\left(R^2  r^2 = 9\right)$ $\displaystyle 2R\frac{dR}{dt}  2r\frac{dr}{dt} = 0$ $\displaystyle R\frac{dR}{dt}  r\frac{dr}{dt} = 0$ Quote:
$\displaystyle \frac{dA_R}{dt} = 2\pi R \frac{dR}{dt} = 10 \pi \implies R\frac{dR}{dt} = 5$ Quote:
$\displaystyle C_r = 2\pi r$ $\displaystyle \frac{dC_r}{dt} = 2\pi \frac{dr}{dt}$ 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 06:42 PM.  
April 15th, 2015, 09:15 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,981 Thanks: 1574  Quote:
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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