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anonymous
 one year ago
Related rates question!!!
A circle is inscribed in a square. The area of the circle is increasing at a constant rate^2/sec. As the circle expands the square expands to keep the circle inscribed at what rate is the area of the square increasing in in^2/sec ?
anonymous
 one year ago
Related rates question!!! A circle is inscribed in a square. The area of the circle is increasing at a constant rate^2/sec. As the circle expands the square expands to keep the circle inscribed at what rate is the area of the square increasing in in^2/sec ?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443498489098:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1is there a typo? `The area of the circle is increasing at a constant rate^2/sec`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep sorry about that " the area of the circle is increasing at a constant rate of 15pi in^2/sec

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1oh ok that makes more sense now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0;P , alright so for the area of the square im getting (2r)^2 , not sure where to go on frem there

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443498822076:dw hopefully you see how the diameter d is equal to the side length of the square

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yes the side length is 2r

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1A = (2r)^2 = 4r^2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1if A = 4r^2, then what is dA/dt ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay but we dont have dr/dt we only have da/dt of the circle

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Circle Ac = pi*r^2 dAc/dt = 2*pi*r*dr/dt 15pi = 2*pi*r*dr/dt

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Rectangle Ar = 4r^2 dAr/dt = 8r*dr/dt try to connect dAc/dt and dAr/dt somehow

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm thinking to divide the two so maybe dAc/dt divided by dAr/dt

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1(dAc/dt)/(dAr/dt) = [ 2*pi*r*dr/dt ] / [8r*dr/dt] the 'r' and 'dr/dt' terms will cancel to get (dAc/dt)/(dAr/dt) = pi/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay one sec let me process this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay nvm i found an easier way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so we know that dr/dt = 15/2r and we know that dAr/dt = 8r dr/dt so if you just plug in dr/dt into the dAr/dt formula you get 120r/2r which gives us 60 which i know to be the answer

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.160 is correct

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1alternatively, you can let x = dAr/dt, so (dAc/dt)/(dAr/dt) = pi/4 (15pi)/(x) = pi/4 15pi*4 = pi*x 60pi = pi*x x*pi = 60pi x = 60

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep didn't see that , thanks for your help really appreciate it :D
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