#$&* course Mth 277 11/19 8 11.5*********************************************
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Given Solution: z = x y + 1 = cos(3t) * cot(3 t) + 1, so dz/dt = (cos(3t)) ' cot(3t) + cos(3t) (cot(3t) ) ' = -3 sin(3t) cot(3t) - cos(3 t) sec^2(3 t), which could be simplified. Using the chain rule dz/dt = dz/dx dx/dt + dz/dy dy/dt = y * (-3 sin(3t) ) + x * (-sec^2(3 t)) = -3 sin(3 t) cot(3 t) + cos(3 t) * (-sec^2(3 t) ), which could also be simplified but is clearly equal to the previous expression. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): in your dz/dt I think you left 3 off of - cos(3t) sec^2(3t) because u = 3t and u' = 3 ------------------------------------------------ Self-critique rating:
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Given Solution: z = x^2 + y^2 = (u cos(v))^2 + (u + v^2) ^ 2 = u^2 cos^2(v) + u^2 + 2 u v^2 + v^4. So z_u = 2 u cos^2(v) + 2 u + 2 v^2 and z_v = -2 u^2 cos(v) sin(v) + 4 u v + 4 v^3. Applying the chain rule: F_x = 2 x and F_y = 2 y. dx/du = cos(v), dx/dv = - u sin(v), dy/du = 1 and dy/dv = 2 v. Thus dz/du = dz/dx * dx/du + dz/dy * dy/du = 2 x * cos(v) * 2 v + 2 y * 1 = 2 u cos(v) * cos(v) + 2 (u + v^2) * 1. When simplified this agrees with our previous result z_u = 2 u cos^2(v) + 2 u + 2 v^2 . dz/dv works out in an analogous manner. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): z_u should have +v^4 on the end z_v should have +u^2 in the middle of that ???????? in your equation for dz/du you have = dz/dx*dx/du at the start. well dz/du i can see is just F_x_u which is 2x. so you fill in 2x*(cos(v)). and thats it. why do you have * 2v in there with it???? there are similarities in our answers but this is a pretty tedious situation
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Given Solution: w_r can be written dw/dr, and we have dw/dr = dw/dx dx/dr + dw/dy dy/dr + dw/dz dz/dr = e^(x - y + 3 z^2) * 1 - e^(x - y + 3 z^2) * 3 + 6 z e^(x - y + 3 z^2) * (r s) cos(r s t). Simplifying and substituting for x, y and z we get (-2 + 6 sin(rst) ) e^(-2 r + 3 t + 3 ( sin^2(rst)) ) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok here i only took the partial derivative of r in the function w. did not include dw/dx and all that jazz. I failed to realize that there is indivual x,y,and z terms here, because i substituted early in for x,y, and z into the origional equation ""masking"" the x,y, and z terms but i see how you got this, it is similar to the past problems and I'm sure i could do this if i worked carefully enough. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. The combined resistance of three resistors R is given by the formula 1/R = 1/(R1) + 1/(R2) + 1/(R3). Suppose that at a certain instant R1 = 150 ohm, R2 = 300 ohm, and R3 = 450 ohm. R1 and R3 are decreasing at a rate of 3 ohm/sec and R2 is increasing at a rate of 4 ohm/sec. How fast is R changing at this instant and is it increasing or decreasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: decreasing by 0.0000025 ohm/s (1/900) - (1/898) = that ^^ confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Regarding R, R1, R2 and R3 as functions of t, we take the t derivative of both sides and get -1/R^2 dR/dt = -1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt so that dR/dt = (-1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt) * R^2 Substituting the given values for the three resistances and the three rates of change we get dR/dt = (-1 / (150 ohms)^2 * (-3 ohms/sec) - 1 / (300 ohms) * (+4 ohms/sec) - 1 / (450 ohms) * (-3 ohms/sec) ) * R^2 = 7 / (67500 ohm sec) * (900/11 ohms) ^2 = .69 ohms / sec, approx.. You can verify that for the given values of R1, R2 and R3 we get R = 900/11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok didn't realize that because it was a function of t that you had to take deriv. but I understand it ******************** end of frs to here snd follows ************ start Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. The combined resistance of three resistors R is given by the formula 1/R = 1/(R1) + 1/(R2) + 1/(R3). Suppose that at a certain instant R1 = 150 ohm, R2 = 300 ohm, and R3 = 450 ohm. R1 and R3 are decreasing at a rate of 3 ohm/sec and R2 is increasing at a rate of 4 ohm/sec. How fast is R changing at this instant and is it increasing or decreasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: decreasing by 0.0000025 ohm/s (1/900) - (1/898) = that ^^ confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Regarding R, R1, R2 and R3 as functions of t, we take the t derivative of both sides and get -1/R^2 dR/dt = -1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt so that dR/dt = (-1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt) * R^2 Substituting the given values for the three resistances and the three rates of change we get dR/dt = (-1 / (150 ohms)^2 * (-3 ohms/sec) - 1 / (300 ohms) * (+4 ohms/sec) - 1 / (450 ohms) * (-3 ohms/sec) ) * R^2 = 7 / (67500 ohm sec) * (900/11 ohms) ^2 = .69 ohms / sec, approx.. You can verify that for the given values of R1, R2 and R3 we get R = 900/11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok didn't realize that because it was a function of t that you had to take deriv. but I understand it ******************** end of frs to here snd follows ************ start Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `q004. The combined resistance of three resistors R is given by the formula 1/R = 1/(R1) + 1/(R2) + 1/(R3). Suppose that at a certain instant R1 = 150 ohm, R2 = 300 ohm, and R3 = 450 ohm. R1 and R3 are decreasing at a rate of 3 ohm/sec and R2 is increasing at a rate of 4 ohm/sec. How fast is R changing at this instant and is it increasing or decreasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: decreasing by 0.0000025 ohm/s (1/900) - (1/898) = that ^^ confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Regarding R, R1, R2 and R3 as functions of t, we take the t derivative of both sides and get -1/R^2 dR/dt = -1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt so that dR/dt = (-1/R1^2 d(R1)/dt -1/R2^2 d(R2)/dt -1/R3^2 d(R3)/dt) * R^2 Substituting the given values for the three resistances and the three rates of change we get dR/dt = (-1 / (150 ohms)^2 * (-3 ohms/sec) - 1 / (300 ohms) * (+4 ohms/sec) - 1 / (450 ohms) * (-3 ohms/sec) ) * R^2 = 7 / (67500 ohm sec) * (900/11 ohms) ^2 = .69 ohms / sec, approx.. You can verify that for the given values of R1, R2 and R3 we get R = 900/11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok didn't realize that because it was a function of t that you had to take deriv. but I understand it ******************** end of frs to here snd follows ************ start Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!