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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #252 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}-\frac{(2x^{3}y^{3} - 3x^{6} + y^{6})}{(2x^{6} + 4y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }3\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{28}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-1\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{1}{4}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{1}{4}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)^{3}y^{3} - 3(0)^{6} + y^{6})}{(2(0)^{6} + 4y^{6})}=\frac{-y^{6}}{4y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{-y^{6}}{4y^{6}}=-\frac{1}{4}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(2x^{3}(x)^{3} - 3x^{6} + (x)^{6})}{(2x^{6} + 4(x)^{6})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #253 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(8x^{2}y^{3} - 16x^{3}y^{2} + 8xy^{4})}{(2x^{5} + 4y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{46}{9}\\&y=x:-\frac{41}{19}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{23}{2}\\&y=x:5\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{36}{13}\\&y=x:-\frac{37}{18}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{2}y^{3} - 16(0)^{3}y^{2} + 8(0)y^{4})}{(2(0)^{5} + 4y^{5})}=\frac{0}{4y^{5}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{4y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(8x^{2}(x)^{3} - 16x^{3}(x)^{2} + 8x(x)^{4})}{(2x^{5} + 4(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #254 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(7x^{2}y - 11x^{3} + 4y^{3})}{(x^{3} + 5y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{48}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }4\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{4}{5}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{4}{65}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{4}{5}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(7(0)^{2}y - 11(0)^{3} + 4y^{3})}{((0)^{3} + 5y^{3})}=\frac{-4y^{3}}{5y^{3}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{-4y^{3}}{5y^{3}}=-\frac{4}{5}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(7x^{2}(x) - 11x^{3} + 4(x)^{3})}{(x^{3} + 5(x)^{3})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #255 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(14x^{3}y - 11x^{4})}{(17xy^{3} + x^{4} + 5y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{60}{23}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{39}{23}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{23}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{46}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{23}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(14(0)^{3}y - 11(0)^{4})}{(17(0)y^{3} + (0)^{4} + 5y^{4})}=\frac{0}{5y^{4}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{5y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(14x^{3}(x) - 11x^{4})}{(17x(x)^{3} + x^{4} + 5(x)^{4})}=-\frac{3}{23}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #256 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}-\frac{(4x^{2}y - 11xy^{2} + 7x^{3})}{(5x^{3} + 4y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{5}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{29}{10}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-32\\&y=x:-\frac{37}{15}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{2}y - 11(0)y^{2} + 7(0)^{3})}{(5(0)^{3} + 4y^{3})}=\frac{0}{4y^{3}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(4x^{2}(x) - 11x(x)^{2} + 7x^{3})}{(5x^{3} + 4(x)^{3})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #257 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}-\frac{(8xy^{4} - 15x^{2}y^{3} + 7x^{5})}{(5x^{5} + y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-9\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{24}{5}\\&y=x:15\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{44}{13}\\&y=x:-\frac{4}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)y^{4} - 15(0)^{2}y^{3} + 7(0)^{5})}{(5(0)^{5} + y^{5})}=\frac{0}{y^{5}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(8x(x)^{4} - 15x^{2}(x)^{3} + 7x^{5})}{(5x^{5} + (x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #258 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(15x^{5}y - 20x^{3}y^{3} + 10x^{6})}{(2x^{6} - x^{4}y^{2} + y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }45\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{95}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{4}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{2}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{5}y - 20(0)^{3}y^{3} + 10(0)^{6})}{(2(0)^{6} - (0)^{4}y^{2} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(15x^{5}(x) - 20x^{3}(x)^{3} + 10x^{6})}{(2x^{6} - x^{4}(x)^{2} + (x)^{6})}=\frac{5}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #259 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(xy^{3} + x^{3}y - 2x^{4})}{(x^{4} + y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{5}{4}\\&y=x:-\frac{43}{19}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{3}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{22}{9}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)y^{3} + (0)^{3}y - 2(0)^{4})}{((0)^{4} + y^{4})}=\frac{0}{y^{4}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(x(x)^{3} + x^{3}(x) - 2x^{4})}{(x^{4} + (x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #260 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}-\frac{(14xy^{5} - 18x^{2}y^{4} + 4x^{6})}{(3x^{6} + 2y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{12}{7}\\&y=x:-41\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{35}{19}\\&y=x:\frac{23}{15}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{35}{16}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(14(0)y^{5} - 18(0)^{2}y^{4} + 4(0)^{6})}{(3(0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(14x(x)^{5} - 18x^{2}(x)^{4} + 4x^{6})}{(3x^{6} + 2(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #261 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(13x^{3}y^{2} + 2x^{5})}{(14x^{2}y^{3} - 10x^{3}y^{2} - 3x^{5} + 15y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{105}{16}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{32}\end{align*}$

$\begin{align*}&\text{y-axis: }15\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{15}{16}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{15}{16}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(13(0)^{3}y^{2} + 2(0)^{5})}{(14(0)^{2}y^{3} - 10(0)^{3}y^{2} - 3(0)^{5} + 15y^{5})}=\frac{0}{-15y^{5}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{-15y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(13x^{3}(x)^{2} + 2x^{5})}{(14x^{2}(x)^{3} - 10x^{3}(x)^{2} - 3x^{5} + 15(x)^{5})}=\frac{15}{16}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #252 : Partial Derivatives

Example Question #253 : Partial Derivatives

Example Question #254 : Partial Derivatives

Example Question #255 : Partial Derivatives

Example Question #256 : Partial Derivatives

Example Question #257 : Partial Derivatives

Example Question #258 : Partial Derivatives

Example Question #259 : Partial Derivatives

Example Question #260 : Partial Derivatives

Example Question #261 : Partial Derivatives

All Calculus 3 Resources

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