Section 14.4 Enhanced Assign

Instructor: Shijun Zheng Course: Spring 2023 - Calculus III (MATH2243-A-B) Student: Jalen Wilson Date: 07/11/24 1. 2 2 For the functions w = − 6x − 8y , x = cos t​, and y = sin t​, express dw dt as a function of​t, both by using the chain rule and by expressing w dw in terms of t and differentiating directly with respect to t. Then evaluate Express dw dt dw dw 2. π 2 . as a function of t. t= dw dt at t = π = π 2 . ​(Type an exact​answer, using radicals as​needed.) 0 2 For the given​functions, ​(a) express dw dt as a function of​t, both by using the chain rule and by expressing w in terms of t and differentiating dw directly with respect to t. Then ​(b) evaluate x 2 w = 5ye − ln z​, x = ln t + 1 ​, y = tan ​(a) dt at t = = − 4 cos t sin t Evaluate dt dt Assignment: Section 14.4 Enhanced Assign dw dt = 10t tan −1 dt −1 at the given value of t. t t​, z = e ​; t = 1 t+4 5π = +4 dt t = 1 2 ​(Type an exact​answer, using π as​needed.) ​(b) dw 3. The equation r(t) = (6 sin t) i + (6 cos t) j + (6t) k is the position of a particle in space at time t. Find the​particle's velocity and acceleration 3π vectors. Then write the​particle's velocity at t = as a product of its speed and direction. 2 The velocity vector is v(t) = 6 cos t The acceleration vector is a(t) = The velocity vector at t = v 3π 2 = 6 2 3π 2 i+ − 6 sin t − 6 sin t i+ j+ − 6 cos t 6 k. j+ 0 k. written as a product of the speed and direction is ​[ 0 i+ ​(Type exact​answers, using radicals as​needed.) 1 2 j+ 1 2 k]. 4. Consider the functions z = 6ex ln y​, x = ln (u cos v)​, and y = u sin v. ∂z ∂z ​(a) Express and as functions of u and v both by using the Chain Rule and by expressing z directly in terms of u and v before ∂u ∂v differentiating. ∂z ∂z π . ​(b) Evaluate and at ​(u,v) = 8, ∂u ∂v 6 ∂z ​(a) Find each partial derivative needed to use the Chain Rule to find ∂z ∂x ∂z ∂y ∂x x 6e ln y = 6e = ∂u ∂y x ∂u y = = ∂u 1 u sin v Express z directly in terms of u and v. z = 6u cos v ln (u sin v) Using either​method, ∂z ∂u = 6 cos v(1 + ln (u sin v)) . ​(Type an expression using u and v as the​variables.) Find each partial derivative needed to use the Chain Rule to find ∂z ∂x ∂z ∂y ∂x x 6e ln y = 6e = ∂v ∂y x ∂v y Using either​method, ∂z ∂v = − tan v = u cos v 2 = − 6u ln (u sin v) sin v + 6u cos v sin v ​(Type an expression using u and v as the​variables.) ​(b) ∂z ∂u ∂z ∂v 8, 8, π 6 π = 3 3 (1 + 2 ln (2)) ​(Type an exact​answer.) 6 = − 48 ln (2) + 72 ​(Type an exact​answer.) . ∂z ∂v . . 5. Draw a dependency​diagram, and write a chain rule formula for dz dp ​, where z = k(r,s)​, r = h(p)​, s = g(p). Choose the correct dependency diagram below. A. B. C. z(k) ∂z dz z ∂z ∂r dk ∂z ∂r ∂s s r Choose the correct chain rule formula. A. dz dp B. dz dp C. dz dp D. dz dp = ∂z dr ∂z ds + ∂r dp ∂s dp = ∂z ds ∂z dr + ∂r dp ∂s dp = ∂z dr ∂z ds − ∂r dp ∂s dp = ∂z dr ∂z ds ∂r dp ∂s dp z = k(r,s) z = k(r,s) ∂z ∂z ∂s ∂r s r dr dp ds p dp ∂s ∂p r(p) s(p) ∂z ∂r ∂p r(r,s) ∂s 6. Draw a dependency​diagram, and write a chain rule formula for ∂w ∂x and ∂w ∂y ​, where w = g(t) and t = k(x,y). Choose the correct dependency diagram below. A. B. t = k(x,y) ∂t t ∂x ∂k w ∂w ∂t ∂t ∂x t ∂y ∂y t y ∂x B. ∂w ∂x C. ∂w ∂x D. ∂w ∂x = ∂t ∂t ∂x ∂y = ∂x ∂y ∂y B. ∂w ∂y C. ∂w ∂y D. ∂w ∂y = = = = ∂x y dx dy dw dw w ∂t ∂x x t dt ∂t ∂y y below. dt ∂y ∂t ∂t − + dw dt dw dt dw ∂t dt ∂x Choose the correct chain rule formula for A. ∂w ∂w dw ∂y x ∂t w = g(t) ∂t dw ∂t = = D. t = k(x,y) ∂x ∂x Choose the correct chain rule formula for A. ∂w ∂t x ∂t ∂w ∂w C. w = g(t) ∂w ∂y below. dw ∂t dt ∂y ∂t ∂t ∂x ∂y − dw dt dw ∂t dt ∂x ∂t ∂t ∂x ∂y + dw dt 7. Assuming that the equation below defines y as a differentiable function of​x, find the value of dy / dx at the given point. 2 2 4x + xy + y − 10 = 0​, (1,2) dy dx (1,2) = −2 ​(Type an integer or a simplified​fraction.)