#$&* course Mth 277 11.5*********************************************
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Given Solution: z = x y + 1 = cos(3t) * cot(3 t) + 1, so dz/dt = (cos(3t)) ' cot(3t) + cos(3t) (cot(3t) ) ' = -3 sin(3t) cot(3t) - cos(3 t) sec^2(3 t), which could be simplified. Using the chain rule dz/dt = dz/dx dx/dt + dz/dy dy/dt = y * (-3 sin(3t) ) + x * (-sec^2(3 t)) = -3 sin(3 t) cot(3 t) + cos(3 t) * (-sec^2(3 t) ), which could also be simplified but is clearly equal to the previous expression. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Let F(x,y) = x^2 + y^2 where x(u,v) = u cos(v) and y(u,v) = u + v^2. Let z = F(x(u,v),y(u,v)). Find z_u and z_v in the following ways. Expressing z explicitly in terms of u and v. Apply the chain rule for two independent parameters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x,y) =x^2 + y^2 x(u,v) =u cos(v) y(u,v) =u + v^2 z=u +v^2 f_u=2u*(cosv)^2+1 and du/dt=cos v f_v=2v-2sin(v)* cos(v)* u^2 and dv/dt=2v [2u(cosv)^2+1*(cosv)] + [2v-2sin(v)*cos(v)*u^2*(2v)] 2u(cosv)^2+(cosv)+2v-4vsin(v)cos(v)u^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. Find w_r where w = e^(x - y + 3z^2) and x = r + t - s, y = 3r - 2t, z = sin(rst). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: w=e^(x-y+3z^2) x=r+t-s y=3r-2t z=sin(rst) w_r=(w_x*x_r)+(w_y*y_r)+(w_z*z_r) w_r=[f_x(e^(x - y + 3z^2))]*[f_r(r+t-s)] + [f_y(e^(x - y + 3z^2))]*[f_r(3r-2t)] + [f_z(e^(x - y + 3z^2))]*[f_r(sin(rst))] w_r=[1e^(x - y + 3z^2)]*[1] + [-e^(x - y + 3z^2)]*[3] + [6z e^(x - y + 3z^2)]*[st*cos (rst)] w_r=[e^(x - y + 3z^2)] + [-3e^(x - y + 3z^2)] + [6z e^(x - y + 3z^2)]*[st*cos (rst)] After simplifying this rather large expression, we can go back and substitue for the variables confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: w_r can be written dw/dr, and we have dw/dr = dw/dx dx/dr + dw/dy dy/dr + dw/dz dz/dr = e^(x - y + 3 z^2) * 1 - e^(x - y + 3 z^2) * 3 + 6 z e^(x - y + 3 z^2) * (r s) cos(r s t). Simplifying and substituting for x, y and z we get (-2 + 6 sin(rst) ) e^(-2 r + 3 t + 3 ( sin^2(rst)) ) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004. The combined resistance of three resistors R is given by the formula 1/R = 1/(R1) + 1/(R2) + 1/(R3). Suppose that at a certain instant R1 = 150 ohm, R2 = 300 ohm, and R3 = 450 ohm. R1 and R3 are decreasing at a rate of 3 ohm/sec and R2 is increasing at a rate of 4 ohm/sec. How fast is R changing at this instant and is it increasing or decreasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/R=1/(R1)+1/(R2)+1/(R3) der -1/R^2(dR/dt)=-1/(R1^2)(dR1/dt)-1/(R2^2)(dR2/dt)-1/(R3^2)(dR3/dt) dR/dt=(-1/(R1^2)(dR1/dt)-1/(R2^2)(dR2/dt)-1/(R3^2)(dR3/dt))(-R^2) dR/dt=((-1/150ohm^2)(-3 ohm/sec)+(-1/300ohm^2)(4 ohm/sec)+(-1/450ohm^2)(-3 ohm/sec))(-R^2) dR/dt=(81.8^2)(.000104) dR/dt=.694 ohms/sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ------------------------------------------------ Self-critique rating:Ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!