Exam 15: Partial Derivatives

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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)=4x2+9y29 at (3,2);R:x+30.1,y+20.1f ( x , y ) = - 4 x ^ { 2 } + 9 y ^ { 2 } - 9 \text { at } ( - 3 , - 2 ) ; R : | x + 3 | \leq 0.1 , | y + 2 | \leq 0.1

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Write a chain rule formula for the following derivative. - wt for w=f(p,q,r);p=g(t),q=h(t),r=k(t)\frac { \partial \mathrm { w } } { \partial \mathrm { t } } \text { for } \mathrm { w } = \mathrm { f } ( \mathrm { p } , \mathrm { q } , \mathrm { r } ) ; \mathrm { p } = \mathrm { g } ( \mathrm { t } ) , \mathrm { q } = \mathrm { h } ( \mathrm { t } ) , \mathrm { r } = \mathrm { k } ( \mathrm { t } )

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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)=xyx2+y2f ( x , y ) = \frac { x y } { x ^ { 2 } + y ^ { 2 } }

(Essay)
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Find the extreme values of the function subject to the given constraint. - f(x,y,z)=4x3y+2z,x2+y2=6zf ( x , y , z ) = 4 x - 3 y + 2 z , \quad x ^ { 2 } + y ^ { 2 } = 6 z

(Multiple Choice)
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Provide an appropriate answer. -Suppose that x2+y2=r2x ^ { 2 } + y ^ { 2 } = r ^ { 2 } and x=rcosθx = r \cos \theta , as in polar coordinates. Find (yθ)r\left( \frac { \partial y } { \partial \theta } \right) _ { r } .

(Multiple Choice)
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Give an appropriate answer. - f(x,y)=exx2+y2f ( x , y ) = \frac { e ^ { - x } } { x ^ { 2 } + y ^ { 2 } }

(Multiple Choice)
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Find the limit. - lim(x,y)(4,3)xlny\lim _ { ( x , y ) \rightarrow ( 4,3 ) } x \ln y

(Multiple Choice)
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At what points is the given function continuous? - f(x,y)=xyx+yf ( x , y ) = \frac { x y } { x + y }

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Use implicit differentiation to find the specified derivative at the given point. -Find yz\frac { \partial y } { \partial z } at the point (2,3,3)( 2,3,3 ) for 2x2+2y2+4z2=0\frac { 2 } { x ^ { 2 } } + \frac { 2 } { y ^ { 2 } } + \frac { 4 } { z ^ { 2 } } = 0 .

(Multiple Choice)
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Sketch a typical level surface for the function. - f(x,y,z)=x2zf ( x , y , z ) = \sqrt { x ^ { 2 } - z }

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Find the absolute maximum and minimum values of the function on the given curve. -Function: f(x,y)=xy;f ( x , y ) = x y ; curve: x236+y281=1,y0\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 81 } = 1 , y \geq 0 . (Use the parametric equations x=6cost,y=9sintx = 6 \cos t , y = 9 \sin t .)

(Multiple Choice)
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Compute the gradient of the function at the given point. - f(x,y)=4x2+6y,(4,3)f ( x , y ) = - 4 x ^ { 2 } + 6 y , \quad ( 4,3 )

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Sketch the surface z = f(x,y). - f(x,y)=4x2y2f ( x , y ) = \sqrt { 4 - x ^ { 2 } - y ^ { 2 } }

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Find the extreme values of the function subject to the given constraint. - f(x,y)=3xy+1,3x2+y2=9f ( x , y ) = 3 x - y + 1 , \quad 3 x ^ { 2 } + y ^ { 2 } = 9

(Multiple Choice)
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Find the domain and range and describe the level curves for the function f(x,y). - f(x,y)=sin1(x2+y2)f ( x , y ) = \sin ^ { - 1 } \left( x ^ { 2 } + y ^ { 2 } \right)

(Multiple Choice)
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Find the specific function value. -Find f(-3, 6) when f(x, y) = 5x + 2y - 3.

(Multiple Choice)
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Find all the second order partial derivatives of the given function. - f(x,y)=ln(x2yx)f ( x , y ) = \ln \left( x ^ { 2 } y - x \right)

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Find the domain and range and describe the level curves for the function f(x,y). - f(x,y)=4x22y2f ( x , y ) = 4 x ^ { 2 } - 2 y ^ { 2 }

(Multiple Choice)
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Solve the problem. -Find the least squares line for the points (-5, -1), (3, -2).

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

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