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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

Example Question #222 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(6xy^{2})}{(16xy^{2} + x^{3} - 4y^{3})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-6\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{6}{169}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{36}{13}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{6}{13}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{6}{13}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(6(0)y^{2})}{(16(0)y^{2} + (0)^{3} - 4y^{3})}=\frac{0}{4y^{3}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(6x(x)^{2})}{(16x(x)^{2} + x^{3} - 4(x)^{3})}=\frac{6}{13}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #221 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(10x^{2}y^{3} + 15x^{4}y)}{(19x^{5} + 8y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{25}{27}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{125}{27}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{250}{27}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{25}{432}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{25}{27}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(10(0)^{2}y^{3} + 15(0)^{4}y)}{(19(0)^{5} + 8y^{5})}=\frac{0}{8y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{8y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(10x^{2}(x)^{3} + 15x^{4}(x))}{(19x^{5} + 8(x)^{5})}=-\frac{25}{27}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1181 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}\frac{(15x^{3}y + x^{4} + 19y^{4})}{(3x^{4} - 13x^{2}y^{2} + 4y^{4})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{19}{60}\\&y=x:-\frac{35}{18}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{175}{3}\\&y=x:-57\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{350}{3}\\&y=x:\frac{57}{2}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{19}{4}\\&y=x:-\frac{35}{6}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }\frac{19}{4}\\&y=x:-\frac{35}{6}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{3}y + (0)^{4} + 19y^{4})}{(3(0)^{4} - 13(0)^{2}y^{2} + 4y^{4})}=\frac{19y^{4}}{4y^{4}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{19y^{4}}{4y^{4}}=\frac{19}{4}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(15x^{3}(x) + x^{4} + 19(x)^{4})}{(3x^{4} - 13x^{2}(x)^{2} + 4(x)^{4})}=-\frac{35}{6}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #225 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{36}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{7}{2}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{1}{2}\\&y=x:0\end{align*}$

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

Correct answer:

$\begin{align*}&\text{x-axis: }-\frac{1}{2}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{3} - 5(0)^{2}y + y^{3})}{(4(0)^{3} + 2y^{3})}=\frac{-y^{3}}{2y^{3}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{-y^{3}}{2y^{3}}=-\frac{1}{2}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(4x^{3} - 5x^{2}(x) + (x)^{3})}{(4x^{3} + 2(x)^{3})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #221 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}\frac{(8x^{6} + 19y^{6})}{(18x^{2}y^{4} - 12x^{4}y^{2} + 5x^{6} + y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{19}{8}\\&y=x:\frac{9}{52}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{27}{2}\\&y=x:-323\end{align*}$

$\begin{align*}&\text{x-axis: }19\\&y=x:\frac{9}{4}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{81}{2}\\&y=x:38\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }19\\&y=x:\frac{9}{4}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(8(0)^{6} + 19y^{6})}{(18(0)^{2}y^{4} - 12(0)^{4}y^{2} + 5(0)^{6} + y^{6})}=\frac{19y^{6}}{y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{19y^{6}}{y^{6}}=19\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(8x^{6} + 19(x)^{6})}{(18x^{2}(x)^{4} - 12x^{4}(x)^{2} + 5x^{6} + (x)^{6})}=\frac{9}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #222 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(16y^{6})}{(2x^{5}y + x^{6} + 6y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{8}{45}\\&y=x:-\frac{2}{9}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{8}{3}\\&y=x:-\frac{16}{9}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{224}{9}\\&y=x:-48\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{176}{9}\\&y=x:40\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }-\frac{8}{3}\\&y=x:-\frac{16}{9}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(16y^{6})}{(2(0)^{5}y + (0)^{6} + 6y^{6})}=\frac{-16y^{6}}{6y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{-16y^{6}}{6y^{6}}=-\frac{8}{3}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(16(x)^{6})}{(2x^{5}(x) + x^{6} + 6(x)^{6})}=-\frac{16}{9}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #228 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{6}{5}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{6}{95}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{42}{5}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{96}{5}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }\frac{6}{5}\\&y=x:0\end{align*}$

Explanation:

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

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Example Question #229 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{5}{12}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{85}{2}\\&y=x:0\end{align*}$

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

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

Correct answer:

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

Explanation:

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

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Example Question #230 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{28}{11}\\&y=x:-46\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-24\end{align*}$

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

$\begin{align*}&\text{x-axis: }-16\\&y=x:\frac{45}{16}\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #1181 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}-\frac{(14xy^{2} - 11y^{3})}{(17xy^{2} - x^{3} + 3y^{3})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{11}{51}\\&y=x:-\frac{3}{361}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{11}{3}\\&y=x:-\frac{3}{19}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{42}{19}\\&y=x:\frac{55}{3}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{6}{19}\\&y=x:-\frac{44}{3}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }\frac{11}{3}\\&y=x:-\frac{3}{19}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(14(0)y^{2} - 11y^{3})}{(17(0)y^{2} - (0)^{3} + 3y^{3})}=\frac{-11y^{3}}{-3y^{3}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{-11y^{3}}{-3y^{3}}=\frac{11}{3}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(14x(x)^{2} - 11(x)^{3})}{(17x(x)^{2} - x^{3} + 3(x)^{3})}=-\frac{3}{19}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #222 : Partial Derivatives

Example Question #221 : Limits

Example Question #1181 : Calculus 3

Example Question #225 : Partial Derivatives

Example Question #221 : Partial Derivatives

Example Question #222 : Partial Derivatives

Example Question #228 : Partial Derivatives

Example Question #229 : Partial Derivatives

Example Question #230 : Partial Derivatives

Example Question #1181 : Calculus 3

All Calculus 3 Resources

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