Exam 13: Partial Differentiation

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Evaluate the limit. Evaluate the limit.    Evaluate the limit.

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Find Find   and   if z = f(x, y) =     sin(x - 2y). and Find   and   if z = f(x, y) =     sin(x - 2y). if z = f(x, y) = Find   and   if z = f(x, y) =     sin(x - 2y). Find   and   if z = f(x, y) =     sin(x - 2y). sin(x - 2y).

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Find Find   if z = sin(uv)cos(uv), where u = e<sup>t</sup> and y = e<sup>2t</sup>. if z = sin(uv)cos(uv), where u = et and y = e2t.

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Let g be a function of a single variable having continuous second order derivatives, and letu(x, y) = g(y + mx) , for some constant real number m. Determine all values of m such that u(x, y) satisfies the partial differential equation Let g be a function of a single variable having continuous second order derivatives, and letu(x, y) = g(y + mx) , for some constant real number m. Determine all values of m such that u(x, y) satisfies the partial differential equation   - 10   + 24   = 0. - 10 Let g be a function of a single variable having continuous second order derivatives, and letu(x, y) = g(y + mx) , for some constant real number m. Determine all values of m such that u(x, y) satisfies the partial differential equation   - 10   + 24   = 0. + 24 Let g be a function of a single variable having continuous second order derivatives, and letu(x, y) = g(y + mx) , for some constant real number m. Determine all values of m such that u(x, y) satisfies the partial differential equation   - 10   + 24   = 0. = 0.

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Describe the level curves of the function f(x,y) = Describe the level curves of the function f(x,y) =   , where a > 0. , where a > 0.

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Find the equation of the normal line to the surface z = Find the equation of the normal line to the surface z =   at the point(2, -2, 1). at the point(2, -2, 1).

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Assume that the relation Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! - 65 + Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! and Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! at (x,y) = (4 , 0). (b) If x = Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! + Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! , y = Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! , find Assume that the relation   - 65 +   = 0 defines z as a differentiable functionof x and y near the point (x , y) = (4 , 0). (a) If z = f(x , y) , find   and   at (x,y) = (4 , 0). (b) If x =   +   , y =   , find   at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule! at (u , v) =(0 , 2)\. Hints :Part (a): First , find the value of z at (x , y) =(4 , 0). Part (b): Use the chain rule!

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The graph below shows level curves f(x,y) = C for a function f and equally spaced values of C. Which of the following functions f is consistent with the graph? (a) f(x,y) = The graph below shows level curves f(x,y) = C for a function f and equally spaced values of C. Which of the following functions f is consistent with the graph? (a) f(x,y) =   +   , (b) f(x,y) =   (c) f(x,y) =    + The graph below shows level curves f(x,y) = C for a function f and equally spaced values of C. Which of the following functions f is consistent with the graph? (a) f(x,y) =   +   , (b) f(x,y) =   (c) f(x,y) =    , (b) f(x,y) = The graph below shows level curves f(x,y) = C for a function f and equally spaced values of C. Which of the following functions f is consistent with the graph? (a) f(x,y) =   +   , (b) f(x,y) =   (c) f(x,y) =    (c) f(x,y) = The graph below shows level curves f(x,y) = C for a function f and equally spaced values of C. Which of the following functions f is consistent with the graph? (a) f(x,y) =   +   , (b) f(x,y) =   (c) f(x,y) =    The graph below shows level curves f(x,y) = C for a function f and equally spaced values of C. Which of the following functions f is consistent with the graph? (a) f(x,y) =   +   , (b) f(x,y) =   (c) f(x,y) =

(Multiple Choice)
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Find the two unit vectors tangent at the point (1, 1, 1) to the curve of intersection of the surfaces xy2 + x2y + z3 = 3 and x3 - y3 - xyz = -1.

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Identify the function f(x,y) whose domain is the shaded region shown in the figure below. Identify the function f(x,y) whose domain is the shaded region shown in the figure below.

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Find an equation of the plane tangent to the surface z = x2 - y2 at the point (2, 1, 3).

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Given z = f(x, y) = Given z = f(x, y) =   y -   , find   . y - Given z = f(x, y) =   y -   , find   . , find Given z = f(x, y) =   y -   , find   . .

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Assuming that the function f has continuous partial derivatives of orders 1 and 2, calculate and simplify Assuming that the function f has continuous partial derivatives of orders 1 and 2, calculate and simplify   f(   y, x   ). f( Assuming that the function f has continuous partial derivatives of orders 1 and 2, calculate and simplify   f(   y, x   ). y, x Assuming that the function f has continuous partial derivatives of orders 1 and 2, calculate and simplify   f(   y, x   ). ).

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Evaluate Evaluate   (3, 2, 1) if f(x, y, z) = x   (yz). (3, 2, 1) if f(x, y, z) = x Evaluate   (3, 2, 1) if f(x, y, z) = x   (yz). (yz).

(Multiple Choice)
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Given that the relation y2 + y Given that the relation y<sup>2</sup> + y   = 14 - sin(xz<sup>2</sup>) +   implicitly defines x as a differentiable function of y and z, find   at the point (0, 3, 4). = 14 - sin(xz2) + Given that the relation y<sup>2</sup> + y   = 14 - sin(xz<sup>2</sup>) +   implicitly defines x as a differentiable function of y and z, find   at the point (0, 3, 4). implicitly defines x as a differentiable function of y and z, find Given that the relation y<sup>2</sup> + y   = 14 - sin(xz<sup>2</sup>) +   implicitly defines x as a differentiable function of y and z, find   at the point (0, 3, 4). at the point (0, 3, 4).

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Show that the function g(x,y) = Show that the function g(x,y) =   is continuous at (x, y) = (0, 0). is continuous at (x, y) = (0, 0).

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Describe the level curves f(x, y) = Describe the level curves f(x, y) =   . .

(Multiple Choice)
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Find the slope of the tangent line to the curve that is the intersection of the surface z = x2 - y2 with the plane x = 2 at the point (2, 1, 3).

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Find an equation of the plane tangent to the surface xy3z2 = 2 at the point (2, 1, -1).

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