course Mth 271 ?U?????????C????K?assignment #021021. `query 21
......!!!!!!!!...................................
09:14:27 **** 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?
......!!!!!!!!...................................
RESPONSE --> x^2 + y^2 = 25 a) I must solve for dy/dt and I am given point (3,4) and dx/dt = 8 First I use implicit differentiation. d/dt[x^2 + y^2] = d/dt[25] 2x dx/dt + 2y dy/dt = 0 2y dy/dt = -2x dx/dt dy/dt = (-x dx/dt) / y Now I substitute the given values dy/dt = -3 * 8 / 4 = -6 b) I must solve for dx/dt and I am given point (4, 3) and dy/dt = -2 d/dt[x^2 + y^2] = d/dt[25] 2x dx/dt + 2y dy/dt = 0 2x dx/dt = -2y dy/dt dx/dt = (-y dy/dt) / x Now I substitue the given values dx/dt = -3 * -2 / 4 = 6/4 = 3/2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:15:06 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:25:11 **** 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?
......!!!!!!!!...................................
RESPONSE --> First, I start with the formula for the volume of a sphere V = (4/3)`pi r^3 Now I differentiate with respect to t d/dt[V] = d/dt [(4/3)`pi r^3] dV/dt = (4/3)`pi (3r^2) dr/dt dV/dt = 4`pi r^2 dr/dt Since r is increasing at 2 inches per minute, we know that dr/dt = 2 a) dr/dt = 2; r = 6 dV/dt = 4`pi * 6^2 * 2 = 288 `pi b) dr/dt = 2; r = 24 dV/dt = 4`pi * 24^2 * 2 = 4608 `pi confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:25:33 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.. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................