2.6+day+3

1.a) Given: dC/dt = 6 in./min. ; as circle expands, square expands to maintain the condition of tangency. Find: dP/dt (P = perimeter of square) A: Since there is no raduis given, and since the square and circle are expanding together, the rate of their change in perimeters is the same, therefore: **dP/dt = 6 in./min.** Correct, Checked by Hayley Linton
 * Worked by Lilly Byrd**


 * 1b. Givens: dc/dt= 6in/min**
 * A=(pie) r^2 of a circle A=lw of a square find the area of both and subtract them to find the area in the middle**
 * r=5**
 * dA/dt=2(pie)r dR/dt dA/dt=dL/dt*w +dW/dt*l**
 * 25(pie)=2(pie)*5 dR/dt diameter=10 so length =10 and width=10**
 * dR/dt=2.5in/min Area=100 100=dl/dt*10+dw/dt*10**
 * worked by: Rachel Rilling**
 * Correct, Checked by Marla Roper**

3B) V=pir^2h+4/3pir^3 dv/dt=2pirdr/dt(dh/dt) + 4pir^2dr/dt 261pi= 2pi(3)(2)dh/dt+ 4pi(9)(2) 261pi+2pi(6)dh/dt +72pi 261pi-72pi/(12pi)= 16cm Worked by Elborz Safarzadeh

Givens: w = 10m ; L = 20m ; dL/dt = 4 m/sec. ; dw/dt = .5 m/sec. d^2 = l^2 + w^2 ; d^2 = (20)^2 + (10)^2 ; d^2 = 500 ; d = 10 √(5) deriv. of pythag.: 2d(dd/dt) = 2w(dw/dt) + dl(dl/dt) plug: (2)(10 √(5))(dd/dt) = (2)(20)(4) + (2)(10)(-.5) simplify: (20 √(5))(dd/dt) = 160 - 10 = 150 150 / (20 √(5)) = (dd/dt) dd/dt = 15/(2 √5) **dd/dt = 3.354 m/sec.** **worked by lilly byrd** **Correct! checked by lilly byrd**
 * 2. b) worked by lilly byrd (because i was supposed to check this one but no one worked it)(12:16 am)**

da/dt=?

 * A =pi(r)^2**

**worked by pascal cuestas**

 * 4. b)**
 * Given: V=12pi r=3 dv/dt=28pi**
 * V=(1/3)pi(r)^2(h)**
 * 12pi=(1/3)pi(r)^2(h)**
 * 36=(3)^2(h)**
 * 4=h**

V=(1/3)pi(r)^2(h) dv/dt=[(2/3)pi(r)(dr/dt)(h)]+(dh/dt)[(1/3)pi(r)^2] 28pi=(2/3)pi(3)(1/2)(4)+(dh/dt)(1/3)pi(3)^2 28pi=4pi+(dh/dt)3pi 24pi=3pi(dh/dt) 8=dh/dt Worked by Marla Roper
 * checked by Rachel Rilling**


 * 4C) ?? **
 * Checked By Elborz Safarzadeh **