course MTH 271 11/10, 1430 assignment #021021. `query 21
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23:47:58 **** Query 2.8.4 dy/dt for (3,4) with x'=8; dx/dt for (4,3) with y'=-2 **** What is dy/dt if x=3, y=4 and dx/dt = 8?
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RESPONSE --> Original equation: x^2 + y^2 = 25 d/dt[x^2] + d/dt[y^2] = d/dt[25] 2x(dx/dt) + 2y(dy/dt) = 0 x=3, y=4, dx/dt=8 2x(dx/dt) + 2y(dy/dt) = 0 2(3)(8) + 2(4)(dy/dt) = 0 48 + 8(dy/dt) = 0 8 (dy/dt) = -48 dy/dt = (-48/8) = -6 x=4, y=3, dy/dt= -2 2x(dx/dt) + 2y(dy/dt) = 0 2(4)(dx/dt) + 2(3)(-2)= 0 8(dx/dt) -12 = 0 8(dx/dt) = 12 (dx/dt) = 12/8 = 3/2 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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23:48:11 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. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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23:50:58 **** 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?
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RESPONSE --> Sphere: V= (4/3) pi r^3 dr/dt = 2 in/min V= 4/3 pi r^3 dV/dt= 4 pi r^2 dr/dt r=6 dV/dt = 4 pi (6^2) (2) dV/dt = 288 pi in/min r=24 dV/dt= 4 pi (24^2) (2) dV/dt= 4608 pi in/min confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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23:53:46 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.. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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