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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 #1281 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}-\frac{(20x^{3}y^{3} - xy^{5})}{(12x^{4}y^{2} + 3x^{6} - 12y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-57\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{19}{39}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{19}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{152}{3}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{19}{3}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(20(0)^{3}y^{3} - (0)y^{5})}{(12(0)^{4}y^{2} + 3(0)^{6} - 12y^{6})}=\frac{0}{-12y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{-12y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(20x^{3}(x)^{3} - x(x)^{5})}{(12x^{4}(x)^{2} + 3x^{6} - 12(x)^{6})}=-\frac{19}{3}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #323 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(7xy^{5} + 4x^{5}y - x^{6})}{(x^{3}y^{3} + 15xy^{5} - 2x^{6} - 5y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{80}{9}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{10}{9}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{2}{27}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{20}{9}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{10}{9}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(7(0)y^{5} + 4(0)^{5}y - (0)^{6})}{((0)^{3}y^{3} + 15(0)y^{5} - 2(0)^{6} - 5y^{6})}=\frac{0}{5y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(7x(x)^{5} + 4x^{5}(x) - x^{6})}{(x^{3}(x)^{3} + 15x(x)^{5} - 2x^{6} - 5(x)^{6})}=-\frac{10}{9}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #324 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{98}{3}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{35}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{7}{30}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{7}{3}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{7}{3}\\&y=x:0\end{align*}$

Explanation:

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

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Example Question #325 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}\frac{(17x^{4}y)}{(23x^{4}y - 5x^{5} + 10y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{17}{28}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{17}{252}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{34}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{323}{28}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{17}{28}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(17(0)^{4}y)}{(23(0)^{4}y - 5(0)^{5} + 10y^{5})}=\frac{0}{-10y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{-10y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(17x^{4}(x))}{(23x^{4}(x) - 5x^{5} + 10(x)^{5})}=\frac{17}{28}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #326 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}\frac{(13y^{4})}{(41x^{2}y^{2} + 2x^{4} + y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

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

$\begin{align*}&\text{y-axis: }\frac{13}{7}\\&y=x:\frac{13}{880}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{39}{22}\\&y=x:-143\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{52}{11}\\&y=x:234\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(13y^{4})}{(41(0)^{2}y^{2} + 2(0)^{4} + y^{4})}=\frac{13y^{4}}{y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{13y^{4}}{y^{4}}=13\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(13(x)^{4})}{(41x^{2}(x)^{2} + 2x^{4} + (x)^{4})}=\frac{13}{44}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #327 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(2xy^{4} + 10x^{4}y - 12x^{5})}{(x^{5} + 2y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{14}{5}\\&y=x:0\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{34}{15}\\&y=x:-\frac{23}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{9}{4}\\&y=x:-\frac{44}{7}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)y^{4} + 10(0)^{4}y - 12(0)^{5})}{((0)^{5} + 2y^{5})}=\frac{0}{2y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{2y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(2x(x)^{4} + 10x^{4}(x) - 12x^{5})}{(x^{5} + 2(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #321 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}\frac{(8x^{4} - 12y^{4})}{(10xy^{3} - 12x^{3}y - 4x^{4} + 8y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }34\\&y=x:\frac{51}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{3}{20}\\&y=x:-\frac{1}{4}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-18\\&y=x:-\frac{9}{2}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(8(0)^{4} - 12y^{4})}{(10(0)y^{3} - 12(0)^{3}y - 4(0)^{4} + 8y^{4})}=\frac{12y^{4}}{-8y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{12y^{4}}{-8y^{4}}=-\frac{3}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(8x^{4} - 12(x)^{4})}{(10x(x)^{3} - 12x^{3}(x) - 4x^{4} + 8(x)^{4})}=-2\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #321 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{43}{7}\\&y=x:-7\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{21}{11}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{49}{10}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #330 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}-\frac{(5x^{3} - 9x^{2}y + 4y^{3})}{(x^{3} + 5y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{56}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{8}{5}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{15}\end{align*}$

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(5(0)^{3} - 9(0)^{2}y + 4y^{3})}{((0)^{3} + 5y^{3})}=\frac{-4y^{3}}{5y^{3}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{-4y^{3}}{5y^{3}}=-\frac{4}{5}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(5x^{3} - 9x^{2}(x) + 4(x)^{3})}{(x^{3} + 5(x)^{3})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #331 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}\frac{(5xy^{2} + 13y^{3})}{(4xy^{2} + 25x^{3} - 2y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{13}{2}\\&y=x:\frac{2}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{1}{2}\\&y=x:\frac{1}{18}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{26}{3}\\&y=x:-52\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{38}{3}\\&y=x:117\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{13}{2}\\&y=x:\frac{2}{3}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(5(0)y^{2} + 13y^{3})}{(4(0)y^{2} + 25(0)^{3} - 2y^{3})}=\frac{-13y^{3}}{2y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{-13y^{3}}{2y^{3}}=-\frac{13}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(5x(x)^{2} + 13(x)^{3})}{(4x(x)^{2} + 25x^{3} - 2(x)^{3})}=\frac{2}{3}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1281 : Calculus 3

Example Question #323 : Partial Derivatives

Example Question #324 : Partial Derivatives

Example Question #325 : Partial Derivatives

Example Question #326 : Partial Derivatives

Example Question #327 : Partial Derivatives

Example Question #321 : Limits

Example Question #321 : Limits

Example Question #330 : Partial Derivatives

Example Question #331 : Limits

All Calculus 3 Resources

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