calulus 3 :

final answers only are okay. Highlight your final answer please

Section 11.5: Problem 1 [1 point} Supposew = E + 3, where y z 2: = 33', y = 2 + sin{5t), and z = 2 + cos[7t]. (Eu: 1. Use the chain rule to nd as a funciion of z, y, z, and t_ Do not rewrite z, y, and z in ten'ns of t, and do not rewrite e3\" as 2. (it Note: Your answer should be an expression in 2:, y, z, and t; es]. "3)( - 4y + 2t" (10 2. USE part '1. to evaluate E when t = U. 5.1 5 3y .12 z Section 11.5: Problem 2 (1 point) Suppose z = x sin(y) , = = 4s'-t', y = -6st. A. Use the chain rule to find Dz and as functions of c, y, s and t. Oz B. Find the numerical values of and Uz when (s, t) = (5, 4). Oz Us (5, 4) Oz (5, 4) =Section 11.5: Problem 3 (1 point) Use the chain rule to find % and " where 7s z = e" tan(y), x = 2s + 2t, y = 3t First the pieces: 2 2 31 75 = 31 dy = 3t dy 75 THE = 3t And putting it all together: Oz Dz dy Oz Ox and DI Us T dy Us THESection 11.5: Problem 4 (1 point) Let w =-5xy -3y2 -02, a = st, y= est Compute (-5, 5) = Ow (-5, 5) = 25y + 5t + 25e -25 at "x + 15g-25 "t - 30y - 10xSection 11.5: Problem 7 (1 point) Consider the surface F(x, y, 2) = a$2 + sin (162) -1 = 0. Find the following partial derivatives 3x Oz 542.87Section 11.5: Problem 9 {1 point] In a simple electric circuit, Ohm's taw states that V = IR, where V is the voltage in Volts, I is the current in Amperes' and R is the resistance in Ohms. Assume that, as the battery wears out, the voltage decreases at 0.02 Volts per second and, as the resistor heats up, the resistance is increasing at I102 Ohms per second, When the resistance is 100 Ohms and the current is 0.04 Amperes, at what rate is the current changing? -Amperes per second Section 11.6: Problem 1 (1 point) Find the directional derivative of f(x, y) = x y* + 2xy at the point (-3, -3) in the direction 0 = 1/4. The gradient of f is: Vf(x, y) = 2xy' + 8x y 3x 212 + 2x+ = (8-'E-)fA 810 405 The directional derivative is: