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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #291 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(32x^{2}y + 10y^{3})}{(6x^{3} - 15x^{2}y + 13y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{10}{13}\\&y=x:-\frac{21}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }42\\&y=x:\frac{60}{13}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{105}{2}\\&y=x:-\frac{140}{13}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{2}{13}\\&y=x:-\frac{21}{10}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{10}{13}\\&y=x:-\frac{21}{2}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(32(0)^{2}y + 10y^{3})}{(6(0)^{3} - 15(0)^{2}y + 13y^{3})}=\frac{-10y^{3}}{13y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{-10y^{3}}{13y^{3}}=-\frac{10}{13}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(32x^{2}(x) + 10(x)^{3})}{(6x^{3} - 15x^{2}(x) + 13(x)^{3})}=-\frac{21}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #292 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(18x^{3}y^{3})}{(x^{5}y - 14xy^{5} - 17x^{3}y^{3} + 4x^{6} + 5y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{3}{14}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{108}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{96}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{6}{7}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{6}{7}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(18(0)^{3}y^{3})}{((0)^{5}y - 14(0)y^{5} - 17(0)^{3}y^{3} + 4(0)^{6} + 5y^{6})}=\frac{0}{5y^{6}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(18x^{3}(x)^{3})}{(x^{5}(x) - 14x(x)^{5} - 17x^{3}(x)^{3} + 4x^{6} + 5(x)^{6})}=\frac{6}{7}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #291 : Partial Derivatives

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

Possible Answers:

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

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

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{14}\end{align*}$

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

Correct answer:

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

Explanation:

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

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Example Question #291 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}\frac{(4x^{4}y - 12y^{5})}{(20x^{2}y^{3} + 14x^{5} + 5y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{24}{13}\\&y=x:\frac{48}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{16}{13}\\&y=x:-\frac{132}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{12}{5}\\&y=x:-\frac{8}{39}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{12}{65}\\&y=x:-\frac{8}{507}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{12}{5}\\&y=x:-\frac{8}{39}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(4(0)^{4}y - 12y^{5})}{(20(0)^{2}y^{3} + 14(0)^{5} + 5y^{5})}=\frac{-12y^{5}}{5y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{-12y^{5}}{5y^{5}}=-\frac{12}{5}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(4x^{4}(x) - 12(x)^{5})}{(20x^{2}(x)^{3} + 14x^{5} + 5(x)^{5})}=-\frac{8}{39}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1251 : Calculus 3

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

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{41}{14}\\&y=x:-39\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{22}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{38}{9}\end{align*}$

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

Correct answer:

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

Explanation:

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

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Example Question #291 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}\frac{(16x^{4}y^{2} - 9xy^{5} + 6y^{6})}{(14x^{4}y^{2} - 18x^{3}y^{3} + 14xy^{5} - 10x^{6} + 2y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

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

$\begin{align*}&\text{y-axis: }-91\\&y=x:-54\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{3}{8}\\&y=x:\frac{13}{4}\end{align*}$

$\begin{align*}&\text{y-axis: }104\\&y=x:42\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(16(0)^{4}y^{2} - 9(0)y^{5} + 6y^{6})}{(14(0)^{4}y^{2} - 18(0)^{3}y^{3} + 14(0)y^{5} - 10(0)^{6} + 2y^{6})}=\frac{6y^{6}}{2y^{6}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{6y^{6}}{2y^{6}}=3\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(16x^{4}(x)^{2} - 9x(x)^{5} + 6(x)^{6})}{(14x^{4}(x)^{2} - 18x^{3}(x)^{3} + 14x(x)^{5} - 10x^{6} + 2(x)^{6})}=\frac{13}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #291 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(16x^{2}y^{3} + 10x^{4}y)}{(5x^{5} - 30xy^{4} + 4y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{104}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{260}{21}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{26}{21}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{13}{147}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{26}{21}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(16(0)^{2}y^{3} + 10(0)^{4}y)}{(5(0)^{5} - 30(0)y^{4} + 4y^{5})}=\frac{0}{-4y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{-4y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(16x^{2}(x)^{3} + 10x^{4}(x))}{(5x^{5} - 30x(x)^{4} + 4(x)^{5})}=\frac{26}{21}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #298 : Partial Derivatives

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

Possible Answers:

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

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{5}{18}\end{align*}$

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

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

Correct answer:

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

Explanation:

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

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Example Question #292 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(13x^{2}y^{3} + 19y^{5})}{(9x^{3}y^{2} + 5x^{5} + 5y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{512}{19}\\&y=x:-\frac{152}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{19}{5}\\&y=x:-\frac{32}{19}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{19}{65}\\&y=x:-\frac{8}{95}\end{align*}$

$\begin{align*}&\text{y-axis: }32\\&y=x:\frac{304}{5}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{19}{5}\\&y=x:-\frac{32}{19}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(13(0)^{2}y^{3} + 19y^{5})}{(9(0)^{3}y^{2} + 5(0)^{5} + 5y^{5})}=\frac{-19y^{5}}{5y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{-19y^{5}}{5y^{5}}=-\frac{19}{5}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(13x^{2}(x)^{3} + 19(x)^{5})}{(9x^{3}(x)^{2} + 5x^{5} + 5(x)^{5})}=-\frac{32}{19}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #292 : Limits

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

Possible Answers:

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

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

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

$\begin{align*}&\text{y-axis: }-\frac{25}{6}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(10(0)^{3})}{(4(0)y^{2} + 15(0)^{2}y - (0)^{3} + 6y^{3})}=\frac{0}{-6y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{-6y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(10x^{3})}{(4x(x)^{2} + 15x^{2}(x) - x^{3} + 6(x)^{3})}=-\frac{5}{12}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #291 : Partial Derivatives

Example Question #292 : Partial Derivatives

Example Question #291 : Partial Derivatives

Example Question #291 : Partial Derivatives

Example Question #1251 : Calculus 3

Example Question #291 : Limits

Example Question #291 : Partial Derivatives

Example Question #298 : Partial Derivatives

Example Question #292 : Partial Derivatives

Example Question #292 : Limits

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