#$&* course Mth 271 11/13 8p *********************************************
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Given Solution: `a At (3,4) you are given dx/dt as x ' = 8. Since 2x dx/dt + 2y dy/dt = 0 we have 2(3) * 8 + 2 * 4 dy/dt = 0 so dy/dt = -48/8 = -6. At (4,3) you are given dy/dt as y' = -2. So you get 2 * 4 dx/dt + 2 * 3 * -2 = 0 so 8 dx/dt - 12 = 0 and therefore 8 dx/dt = 12. Solving for dx/dt we get dx/dt = 12/8 = 3/2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Since 2x dx/dt + 2y dy/dt = 0 - how do we know this?
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Given Solution: `a The shape is a sphere. The volume of a sphere, in terms of its radius, is V = 4/3 `pi r^3. Taking the derivative with respect to t, noting that r is the only variable, we obtain dV/dt = ( 4 `pi r^2) dr/dt You know that r increases at a rate of 2 in / min, which means that dr/dt = 2. Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx. Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n/a Self-critique Rating 3 ********************************************* Question: `q **** problem 2 7th edition Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V =(4/3) pi r^3 dV/dt = ? dr/dt = 2 2 = (1/(4 pi r^2)) (dV/dt) 2 = 1/(4pi (6)^2 (dV/dt) 2 = 1/144pi(dV/dt) dV/dt = 288 pi V = (4/3) pi r^3 dV/dt = ? dr/dt = 2 2 = 1/(4 pi(24)^2) dV/dt 2 = 1/2304pi dV/dt 4608 pi = dV/dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The shape is a sphere. The volume of a sphere, in terms of its radius, is V = 4/3 `pi r^3. Taking the derivative with respect to t, noting that r is the only variable, we obtain dV/dt = ( 4 `pi r^2) dr/dt You know that r increases at a rate of 2 in / min, which means that dr/dt = 2. Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx. Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n/a Self-critique Rating 3 #*&! ********************************************* Question: `q **** problem 2 7th edition Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V =(4/3) pi r^3 dV/dt = ? dr/dt = 2 2 = (1/(4 pi r^2)) (dV/dt) 2 = 1/(4pi (6)^2 (dV/dt) 2 = 1/144pi(dV/dt) dV/dt = 288 pi V = (4/3) pi r^3 dV/dt = ? dr/dt = 2 2 = 1/(4 pi(24)^2) dV/dt 2 = 1/2304pi dV/dt 4608 pi = dV/dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The shape is a sphere. The volume of a sphere, in terms of its radius, is V = 4/3 `pi r^3. Taking the derivative with respect to t, noting that r is the only variable, we obtain dV/dt = ( 4 `pi r^2) dr/dt You know that r increases at a rate of 2 in / min, which means that dr/dt = 2. Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx. Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): n/a Self-critique Rating 3 #*&!#*&!