Question 1

Find the absolute maxima and minima of the function on the given domain.
-f(x, y) = 4x + 3y on the trapezoidal region with vertices (0, 0), (1, 0), (0, 2), and (1, 1)

Accepted Answer

A)  Absolute maximum: 6 at (0, 2); absolute minimum: 0 at (0, 0) 
B)  Absolute maximum: 7 at (1, 1); absolute minimum: 0 at (0, 0) 
C)  Absolute maximum: 8 at (2, 0); absolute minimum: 4 at (1, 0) 
D)  Absolute maximum: 7 at (1, 1); absolute minimum: 4 at (1, 0) 
A)  Absolute maximum: 6 at (0, 2); absolute minimum: 0 at (0, 0) 
B)  Absolute maximum: 7 at (1, 1); absolute minimum: 0 at (0, 0) 
C)  Absolute maximum: 8 at (2, 0); absolute minimum: 4 at (1, 0) 
D)  Absolute maximum: 7 at (1, 1); absolute minimum: 4 at (1, 0)

Question 2

Sketch a typical level surface for the function.
-$$f ( x , y , z ) = \sqrt { y - x ^ { 2 } - z ^ { 2 } }$$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 3

Solve the problem.
-The resistance $R$ produced by wiring resistors of $R _ { 1 } , R _ { 2 }$, and $R _ { 3 }$ ohms in parallel can be calculated from the formula $\frac { 1 } { R } = \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 2 } } + \frac { 1 } { R _ { 3 } } .$
If $\mathrm { R } _ { 1 } , \mathrm { R } _ { 2 }$, and $\mathrm { R } _ { 3 }$ are measured to be 4 ohms, 7 ohms, and 9 ohms respectively, and if these measurements are accurate to within $0.05$ ohms, estimate the maximum percentage error in computing R.

Accepted Answer

A)  1.13% 
B)  0.95% 
C)  0.76% 
D)  0.57% 
A)  1.13% 
B)  0.95% 
C)  0.76% 
D)  0.57%

Question 4

Solve the problem.
-The table below summarizes the construction cost of a set of homes (excluding the lot cost) along with the square footage of the home's floor space. Find a linear equation that relates the construction cost in thousands of dollars to the floor space in hundreds of square feet by finding the least squares line for the data.

Accepted Answer

A)  y = 8.13x - 12.79 
B)  y = 7.87x - 13.08 
C)  y = 8.13x 
D)  y = 7.97x - 13.54 
A)  y = 8.13x - 12.79 
B)  y = 7.87x - 13.08 
C)  y = 8.13x 
D)  y = 7.97x - 13.54

Question 5

Solve the problem.
-Write an equation for the tangent line to the curve xy = 40 at the point (8, 5).

Accepted Answer

A)  8x + 5y = 40 
B)  8x + 5y = 80 
C)  5x + 8y = 80 
D)  5x + 8y = 40 
A)  8x + 5y = 40 
B)  8x + 5y = 80 
C)  5x + 8y = 80 
D)  5x + 8y = 40

Question 6

Solve the problem.
-$\text { Find parametric equations for the normal line to the surface } x ^ { 2 } - 8 x y z + y ^ { 2 } = - 6 z ^ { 2 } \text { at the point } ( 1,1,1 ) \text {. }$

Accepted Answer

A)  x = -6t + 1, y = -6t + 1, z = 4t + 1 
B)  x = t - 6, y = t - 6, z = t + 4 
C)  x = t + 6, y = t + 6, z = t - 4 
D)  x = -6t + 1, y = 6t + 1, z = 4t + 1 
A)  x = -6t + 1, y = -6t + 1, z = 4t + 1 
B)  x = t - 6, y = t - 6, z = t + 4 
C)  x = t + 6, y = t + 6, z = t - 4 
D)  x = -6t + 1, y = 6t + 1, z = 4t + 1

Question 7

Match the surface show below to the graph of its level curves.
-

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 8

Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0).
-If $f \left( x _ { 0 } , y _ { 0 } \right) = - 2$ and the limit of $f ( x , y )$ exists as $( x , y )$ approaches $\left( x _ { 0 } , y _ { 0 } \right)$, what can you say about the continuity of $\mathrm { f } ( \mathrm { x } , \mathrm { y } )$ at the point $\left( \mathrm { x } _ { 0 } , \mathrm { y } _ { 0 } \right)$ ? Give reasons for your answer.

Accepted Answer

The answer of Find two paths of approach from which...

Question 9

Find the limit.
-$$\lim _ { ( x , y ) \rightarrow ( 0,1 ) } \frac { y ^ { 5 } \sin x } { x }$$

Accepted Answer

A)  $\infty$ 
B)  1 
C)  0 
D)  No limit 
A)  $\infty$ 
B)  1 
C)  0 
D)  No limit

Question 10

Use Taylor's formula to find the requested approximation of f(x, y) near the origin.
-Quadratic approximation to f(x, y) = sin(10x + y)

Accepted Answer

A)  $10 x + y + 5 x ^ { 2 } + \frac { 1 } { 2 } y ^ { 2 }$ 
B)  $10 x + y + 5 x ^ { 2 } + \frac { 1 } { 2 } x y + \frac { 1 } { 2 } y ^ { 2 }$ 
C)  $10 x + y$ 
D)  $10 x + y + 5 x ^ { 2 } + y ^ { 2 }$ 
A)  $10 x + y + 5 x ^ { 2 } + \frac { 1 } { 2 } y ^ { 2 }$ 
B)  $10 x + y + 5 x ^ { 2 } + \frac { 1 } { 2 } x y + \frac { 1 } { 2 } y ^ { 2 }$ 
C)  $10 x + y$ 
D)  $10 x + y + 5 x ^ { 2 } + y ^ { 2 }$

Question 11

Answer the question.
-Describe the results of applying the method of Lagrange multipliers to a function f(x, y) if the points (x, y) are constrained to follow a curve g(x, y) = c that is everywhere perpendicular to the level curves of f. Assume that
Both f(x, y) and g(x, y) satisfy all the requirements and conditions for the method to be applicable.

Accepted Answer

A)  Generally, local extrema of f(x, y) occur at points on the curve g(x, y) = c where the curve becomes tangent to a level curve of f(x, y). Since the curve defined by g(x, y) = c is everywhere perpendicular to the level
Curves of f(x, y) for this particular case, it is never tangent to a level curve, and there are no local extrema
Along g(x,y) = c. The method of Lagrange multipliers will fail to find any local extrema since there are
None. 
B)  The results cannot be generally predicted. Specific expressions for $\mathrm { f } ( \mathrm { x } , \mathrm { y } )$ and $g ( \mathrm { x } , \mathrm { y } ) = \mathrm { c }$ are required. 
C)  Since $\bar { V }$ is everywhere parallel to $\nabla g$, there will be a single local minimum and a single local maximum along $g ( x , y ) = c$. Applying the method of Lagrange multipliers should identify the locations of these two local extrema. 
D)  Since $\nabla$ is everywhere parallel to $\nabla g$, every point on $g ( x , y ) = c$ is a local extremum. Applying the method of Lagrange multipliers should yield the equation $g ( x , y ) = c$. 
A)  Generally, local extrema of f(x, y) occur at points on the curve g(x, y) = c where the curve becomes tangent to a level curve of f(x, y). Since the curve defined by g(x, y) = c is everywhere perpendicular to the level
Curves of f(x, y) for this particular case, it is never tangent to a level curve, and there are no local extrema
Along g(x,y) = c. The method of Lagrange multipliers will fail to find any local extrema since there are
None. 
B)  The results cannot be generally predicted. Specific expressions for $\mathrm { f } ( \mathrm { x } , \mathrm { y } )$ and $g ( \mathrm { x } , \mathrm { y } ) = \mathrm { c }$ are required. 
C)  Since $\bar { V }$ is everywhere parallel to $\nabla g$, there will be a single local minimum and a single local maximum along $g ( x , y ) = c$. Applying the method of Lagrange multipliers should identify the locations of these two local extrema. 
D)  Since $\nabla$ is everywhere parallel to $\nabla g$, every point on $g ( x , y ) = c$ is a local extremum. Applying the method of Lagrange multipliers should yield the equation $g ( x , y ) = c$.

Question 12

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0).
-$$f ( x , y ) = \frac { 9 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$$

Accepted Answer

A)  $- 1$ 
B)  0 
C)  1 
D)  $\pi$ 
A)  $- 1$ 
B)  0 
C)  1 
D)  $\pi$

Question 13

Match the surface show below to the graph of its level curves.
-

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 14

Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0).
-$$f ( x , y ) = \frac { | x y | } { x y }$$

Accepted Answer

Answers will vary. O

Question 15

Find an upper bound for the magnitude |E| of the error in the approximation f(x, y) ≈ L(x, y) at the given point over the
given region R.
-$$f ( x , y , z ) = \ln ( 2 x + 4 y + 2 z ) \text { at } ( 1,1,1 ) ; R : | x - 1 | \leq 0.1 , | y - 1 | \leq 0.1 , | z - 1 | \leq 0.1$$

Accepted Answer

A)  $| E | \leq 0.0167$ 
B)  $| E | \leq 0.0156$ 
C)  $| E | \leq 0.0139$ 
D)  $| \mathrm { E } | \leq 0.0208$ 
A)  $| E | \leq 0.0167$ 
B)  $| E | \leq 0.0156$ 
C)  $| E | \leq 0.0139$ 
D)  $| \mathrm { E } | \leq 0.0208$

Question 16

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0).
-$$f ( x , y ) = \frac { x ^ { 2 } y + x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } }$$

Accepted Answer

A)  0 
B)  2 
C)  1 
D)  No limit 
A)  0 
B)  2 
C)  1 
D)  No limit

Question 17

Solve the problem.
-Find the distance between the skew lines x = t - 6, y = t, z = 2t And X = t, y = t, z = -t.

Accepted Answer

A)  $2 \sqrt { 3 }$ 
B)  $2 \sqrt { 2 }$ 
C)  $3 \sqrt { 2 }$ 
D)  $3 \sqrt { 3 }$ 
A)  $2 \sqrt { 3 }$ 
B)  $2 \sqrt { 2 }$ 
C)  $3 \sqrt { 2 }$ 
D)  $3 \sqrt { 3 }$

Question 18

Find the equation for the level surface of the function through the given point.
-$$f ( x , y , z ) = \sum _ { n = 0 } ^ { \infty } \frac { ( z y + x ) ^ { n } } { 2 ^ { n } n _ { x } ^ { n } } , ( 6,4,4 )$$

Accepted Answer

A)  $\frac { z y + x } { 2 x } = \frac { 11 } { 6 }$ 
B)  $\frac { ( z y + x ) ^ { n } } { 2 n _ { x ^ { n } } } = \frac { 11 } { 6 }$ 
C)  The level surfaces cannot be determined. 
D)  $\sum _ { n = 0 } ^ { \infty } \frac { ( z y + x ) ^ { n } } { 2 ^ { n } x ^ { n } } = \frac { 11 } { 6 }$ 
A)  $\frac { z y + x } { 2 x } = \frac { 11 } { 6 }$ 
B)  $\frac { ( z y + x ) ^ { n } } { 2 n _ { x ^ { n } } } = \frac { 11 } { 6 }$ 
C)  The level surfaces cannot be determined. 
D)  $\sum _ { n = 0 } ^ { \infty } \frac { ( z y + x ) ^ { n } } { 2 ^ { n } x ^ { n } } = \frac { 11 } { 6 }$

Question 19

Solve the problem.
-The radius r and height h of a cylinder are changing with time. At the instant in question, r = 2 cm, h = 6 cm, dr/dt = 0.03 cm/sec and dh/dt = -0.03 cm/sec. At what rate is the cylinder's volume changing at that instant?

Accepted Answer

A)  $2.64 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$ 
B)  $0.75 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$ 
C)  $0.94 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$ 
D)  $1.88 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$ 
A)  $2.64 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$ 
B)  $0.75 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$ 
C)  $0.94 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$ 
D)  $1.88 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$

Question 20

Find all the second order partial derivatives of the given function.
-$f ( x , y ) = \cos \left( x y ^ { 2 } \right)$

Accepted Answer

A)  $\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - y ^ { 4 } \cos x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = - 2 x \left[ 2 x y ^ { 2 } \cos \left( x y ^ { 2 } \right) + \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } =$ $- 2 y \left[ x y ^ { 2 } \cos \left( x y ^ { 2 } \right) + \sin \left( x y ^ { 2 } \right) \right] ;$ 
B)  $\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = y ^ { 2 } \sin x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = 2 \left[ 2 y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } = 2 y \left[ y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right]$ 
C)  $\frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { x } ^ { 2 } } = - \mathrm { y } ^ { 2 } \sin x \mathrm { y } ^ { 2 } ; \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { y } ^ { 2 } } = 2 \left[ \sin \left( \mathrm { xy } ^ { 2 } \right) - 2 \mathrm { y } ^ { 2 } \cos \left( x \mathrm { y } ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { y } \partial \mathrm { x } } = \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { x } \partial \mathrm { y } } = 2 \mathrm { y } \left[ \sin \left( x \mathrm { y } ^ { 2 } \right) - \mathrm { y } ^ { 2 } \cos \left( x \mathrm { y } ^ { 2 } \right) \right]$ 
D)  $\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - y ^ { 2 } \sin x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = 2 y \left[ 2 y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } = 2 \left[ y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right]$ 
A)  $\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - y ^ { 4 } \cos x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = - 2 x \left[ 2 x y ^ { 2 } \cos \left( x y ^ { 2 } \right) + \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } =$ $- 2 y \left[ x y ^ { 2 } \cos \left( x y ^ { 2 } \right) + \sin \left( x y ^ { 2 } \right) \right] ;$ 
B)  $\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = y ^ { 2 } \sin x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = 2 \left[ 2 y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } = 2 y \left[ y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right]$ 
C)  $\frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { x } ^ { 2 } } = - \mathrm { y } ^ { 2 } \sin x \mathrm { y } ^ { 2 } ; \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { y } ^ { 2 } } = 2 \left[ \sin \left( \mathrm { xy } ^ { 2 } \right) - 2 \mathrm { y } ^ { 2 } \cos \left( x \mathrm { y } ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { y } \partial \mathrm { x } } = \frac { \partial ^ { 2 } \mathrm { f } } { \partial \mathrm { x } \partial \mathrm { y } } = 2 \mathrm { y } \left[ \sin \left( x \mathrm { y } ^ { 2 } \right) - \mathrm { y } ^ { 2 } \cos \left( x \mathrm { y } ^ { 2 } \right) \right]$ 
D)  $\frac { \partial ^ { 2 } f } { \partial x ^ { 2 } } = - y ^ { 2 } \sin x y ^ { 2 } ; \frac { \partial ^ { 2 } f } { \partial y ^ { 2 } } = 2 y \left[ 2 y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right] ; \frac { \partial ^ { 2 } f } { \partial y \partial x } = \frac { \partial ^ { 2 } f } { \partial x \partial y } = 2 \left[ y ^ { 2 } \cos \left( x y ^ { 2 } \right) - \sin \left( x y ^ { 2 } \right) \right]$

Find the absolute maxima and minima of the function on the given domain. -f(x, y) = 4x + 3y on the trapezoidal region with vertices (0, 0), (1, 0), (0, 2), and (1, 1)

Sketch a typical level surface for the function. - $f ( x , y , z ) = \sqrt { y - x ^ { 2 } - z ^ { 2 } }$

Solve the problem. -Write an equation for the tangent line to the curve xy = 40 at the point (8, 5).

Solve the problem. - $\text { Find parametric equations for the normal line to the surface } x ^ { 2 } - 8 x y z + y ^ { 2 } = - 6 z ^ { 2 } \text { at the point } ( 1,1,1 ) \text {. }$

Match the surface show below to the graph of its level curves. -

Find the limit. - $\lim _ { ( x , y ) \rightarrow ( 0,1 ) } \frac { y ^ { 5 } \sin x } { x }$

Use Taylor's formula to find the requested approximation of f(x, y) near the origin. -Quadratic approximation to f(x, y) = sin(10x + y)

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0). - $f ( x , y ) = \frac { 9 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$

Match the surface show below to the graph of its level curves. -

Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0). - $f ( x , y ) = \frac { | x y | } { x y }$

Find an upper bound for the magnitude |E| of the error in the approximation f(x, y) ≈ L(x, y) at the given point over the given region R. - $f ( x , y , z ) = \ln ( 2 x + 4 y + 2 z ) \text { at } ( 1,1,1 ) ; R : | x - 1 | \leq 0.1 , | y - 1 | \leq 0.1 , | z - 1 | \leq 0.1$

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0). - $f ( x , y ) = \frac { x ^ { 2 } y + x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } }$

Solve the problem. -Find the distance between the skew lines x = t - 6, y = t, z = 2t And X = t, y = t, z = -t.

Find the equation for the level surface of the function through the given point. - $f ( x , y , z ) = \sum _ { n = 0 } ^ { \infty } \frac { ( z y + x ) ^ { n } } { 2 ^ { n } n _ { x } ^ { n } } , ( 6,4,4 )$

Solve the problem. -The radius r and height h of a cylinder are changing with time. At the instant in question, r = 2 cm, h = 6 cm, dr/dt = 0.03 cm/sec and dh/dt = -0.03 cm/sec. At what rate is the cylinder's volume changing at that instant?

Find all the second order partial derivatives of the given function. - $f ( x , y ) = \cos \left( x y ^ { 2 } \right)$

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Exam 15: Partial Derivatives

Find the absolute maxima and minima of the function on the given domain. -f(x, y) = 4x + 3y on the trapezoidal region with vertices (0, 0), (1, 0), (0, 2), and (1, 1)

Sketch a typical level surface for the function. - f(x,y,z)=y−x2−z2f ( x , y , z ) = \sqrt { y - x ^ { 2 } - z ^ { 2 } }f(x,y,z)=y−x2−z2​

Solve the problem. -Write an equation for the tangent line to the curve xy = 40 at the point (8, 5).

Match the surface show below to the graph of its level curves. -

Find the limit. - lim⁡(x,y)→(0,1)y5sin⁡xx\lim _ { ( x , y ) \rightarrow ( 0,1 ) } \frac { y ^ { 5 } \sin x } { x }lim(x,y)→(0,1)​xy5sinx​

Use Taylor's formula to find the requested approximation of f(x, y) near the origin. -Quadratic approximation to f(x, y) = sin(10x + y)

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0). - f(x,y)=9xyx2+y2f ( x , y ) = \frac { 9 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }f(x,y)=x2+y2​9xy​

Match the surface show below to the graph of its level curves. -

Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0). - f(x,y)=∣xy∣xyf ( x , y ) = \frac { | x y | } { x y }f(x,y)=xy∣xy∣​

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0). - f(x,y)=x2y+xy2x2+y2f ( x , y ) = \frac { x ^ { 2 } y + x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } }f(x,y)=x2+y2x2y+xy2​

Solve the problem. -Find the distance between the skew lines x = t - 6, y = t, z = 2t And X = t, y = t, z = -t.

Find the equation for the level surface of the function through the given point. - f(x,y,z)=∑n=0∞(zy+x)n2nnxn,(6,4,4)f ( x , y , z ) = \sum _ { n = 0 } ^ { \infty } \frac { ( z y + x ) ^ { n } } { 2 ^ { n } n _ { x } ^ { n } } , ( 6,4,4 )f(x,y,z)=∑n=0∞​2nnxn​(zy+x)n​,(6,4,4)

Solve the problem. -The radius r and height h of a cylinder are changing with time. At the instant in question, r = 2 cm, h = 6 cm, dr/dt = 0.03 cm/sec and dh/dt = -0.03 cm/sec. At what rate is the cylinder's volume changing at that instant?

Find all the second order partial derivatives of the given function. - f(x,y)=cos⁡(xy2)f ( x , y ) = \cos \left( x y ^ { 2 } \right)f(x,y)=cos(xy2)

Functions

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Integrals and Transcendental Functions

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Infinite Sequences and Series

Parametric Equations and Polar Coordinates

Vectors and the Geometry of Space

Vector-Valued Functions and Motion in Space

Multiple Integrals

Integrals and Vector Fields

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Sketch a typical level surface for the function. - $f ( x , y , z ) = \sqrt { y - x ^ { 2 } - z ^ { 2 } }$

Find the limit. - $\lim _ { ( x , y ) \rightarrow ( 0,1 ) } \frac { y ^ { 5 } \sin x } { x }$

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0). - $f ( x , y ) = \frac { 9 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$

Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0). - $f ( x , y ) = \frac { | x y | } { x y }$

Use polar coordinates to find the limit of the function as (x, y) approaches (0, 0). - $f ( x , y ) = \frac { x ^ { 2 } y + x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } }$

Find the equation for the level surface of the function through the given point. - $f ( x , y , z ) = \sum _ { n = 0 } ^ { \infty } \frac { ( z y + x ) ^ { n } } { 2 ^ { n } n _ { x } ^ { n } } , ( 6,4,4 )$

Find all the second order partial derivatives of the given function. - $f ( x , y ) = \cos \left( x y ^ { 2 } \right)$