Homework #8 (7092647) Current Score: Due: Wed Mar 25 2015 11:59 PM EDT 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 0/12 0/1 0/8 0/1 0/2 0/8 0/2 0/3 0/3 0/2 0/4 0/8 0/1 0/6 0/1 0/2 0/2 0/3 0/6 Question Points 1. 0/75 Total 0/75 0/12 points SCalcET7 14.2.013.MI.SA. [3159505] - This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Find the limit, if it exists. 2. 0/1 points SCalcET7 14.2.038. [2004958] Determine the set of points at which the function is continuous. {(x, y) | (x, y) ≠ (0, 0)} {(x, y) | x and y ≠ 0} {(x, y) | x > 0 and y > 0} {(x, y) | x · y ≠ 0} {(x, y) | x and y } - 3. 0/8 points SCalcET7 14.2.AE.003. [2005157] Video Example EXAMPLE 3 SOLUTION If f(x, y) = 6xy2 7x2 + y4 , does lim (x, y) → (0, 0) - f(x, y) exist? Let's try to save some time by letting (x, y) → (0, 0) along any nonvertical line through the origin. Then y = mx, where m is the slope, and f(x, y) = f(x, mx) = 2 6x 7x2 + (mx)4 = = 7x2 + m4x4 So f(x, y) → . 7 + m4x2 as (x, y) → (0, 0) along y = mx. Thus f has the same limiting value along every nonvertical line through the origin. But that does not show that the given limit is 0, for if we now let (x, y) → (0, 0) along the parabola x = y2, we have f(x, y) = f(y2, y) · y2 = 7(y2)2 + y4 = 8y4 = So f(x, y) → . as (x, y) → (0, 0) along x = y2. Since different paths lead to different limiting values, the given limit does not exist. 4. 0/1 points SCalcET7 14.2.018. [1887003] Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim xy4 (x, y) → (0, 0) x2 + y8 - 5. 0/2 points SCalcET7 14.3.006. [1888966] - SCalcET7 14.3.015.MI.SA. [1724065] - Determine the signs of the partial derivatives for the function f whose graph is shown below. (a) fx(−x0, y0) positive negative (b) fy(−x0, y0) positive negative 6. 0/8 points This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Find the first partial derivatives of the function. f(x, y) = y9 − 8xy 7. 0/2 points SCalcET7 14.3.019. [1898048] - SCalcET7 14.3.031. [1898101] - SCalcET7 14.3.034. [1898116] - SCalcET7 14.3.047. [1857244] - Find the first partial derivatives of the function. z = (9x + 2y)6 ∂z = ∂x ∂z = ∂y 8. 0/3 points Find the first partial derivatives of the function. f(x, y, z) = xz − 3x5y7z5 fx(x, y, z) = fy(x, y, z) = fz(x, y, z) = 9. 0/3 points Find the first partial derivatives of the function. w = 3zexyz ∂w = ∂x ∂w = ∂y ∂w = ∂z 10. 0/2 points Use implicit differentiation to find ∂z/∂x and ∂z/∂y. x2 + 4y2 + 9z2 = 9 ∂z = ∂x ∂z = ∂y 11. 0/4 points SCalcET7 14.3.053. [1905186] - SCalcET7 14.4.002.MI.SA. [1723990] - Find all the second partial derivatives. f(x, y) = x8y6 + 6x6y fxx(x, y) = fxy(x, y) = fyx(x, y) = fyy(x, y) = 12. 0/8 points This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Find an equation of the tangent plane to the given surface at the specified point. z = 3(x - 1)2 + 2(y + 3)2 + 7, (4, 3, 106) 13. 0/1 points SCalcET7 14.4.006. [1853508] - SCalcET7 14.4.011. [1853638] - Find an equation of the tangent plane to the given surface at the specified point. z = ln(x − 5y), 14. (11, 2, 0) 0/6 points Explain why the function is differentiable at the given point. f(x, y) = 9 + x ln(xy − 7), (2, 4) The partial derivatives are fx(x, y) = fy(2, 4) = . Both fx and fy are continuous functions for xy > Find the linearization L(x, y) of f(x, y) at (2, 4). L(x, y) = and fy(x, y) = , so fx(2, 4) = and and f is differentiable at (2, 4). 15. 0/1 points SCalcET7 14.5.013. [2476110] - SCalcET7 14.5.014. [1905108] - SCalcET7 14.5.015. [1905154] - If z = f(x, y), where f is differentiable, and x g(5) g' (5) fx(3, 0) = g(t) = 3 = −9 y h(5) h' (5) fy(3, 0) = −2 = h(t) = 0 = 9 = 7 find dz/dt when t = 5. dz = dt 16. 0/2 points Let W(s, t) = F(u(s, t), v(s, t)), where F, u, and v are differentiable, and the following applies. u(3, −8) = 5 us(3, −8) = −4 ut(3, −8) = 2 Fu(5, 4) = 9 v(3, −8) = 4 vs(3, −8) = 1 vt(3, −8) = −7 Fv(5, 4) = 0 Find Ws(3, −8) and Wt(3, −8). Ws(3, −8) = Wt(3, −8) = 17. 0/2 points Suppose f is a differentiable function of x and y, and g(u, v) = f(eu + sin v, eu + cos v). Use the table of values to calculate gu(0, 0) and gv(0, 0). f g fx fy (0, 0) 0 3 1 2 (1, 2) 3 0 5 4 gu(0, 0) = gv(0, 0) = 18. 0/3 points SCalcET7 14.5.021. [1853589] Use the Chain Rule to find the indicated partial derivatives. z = x4 + x2y, ∂z , ∂z , ∂z ∂s ∂t ∂u ∂z = ∂s ∂z = ∂t ∂z = ∂u x = s + 2t − u, y = stu2; when s = 5, t = 2, u = 1 - 19. 0/6 points SCalcET7 14.5.038.MI.SA. [1724146] This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise The radius of a right circular cone is increasing at a rate of 1.7 in/s while its height is decreasing at a rate of 2.4 in/s. At what rate is the volume of the cone changing when the radius is 165 in. and the height is 173 in.? Assignment Details -
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