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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #281 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{50}{19}\\&y=x:-\frac{24}{19}\end{align*}$

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

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

$\begin{align*}&\text{y-axis: }-18\\&y=x:-\frac{29}{18}\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #1241 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}\frac{(9x^{3})}{(3x^{3} + 23y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{27}{26}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{54}{13}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{9}{26}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{9}{130}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{9}{26}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(9(0)^{3})}{(3(0)^{3} + 23y^{3})}=\frac{0}{23y^{3}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{23y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(9x^{3})}{(3x^{3} + 23(x)^{3})}=\frac{9}{26}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #281 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}-\frac{(18x^{3}y^{3} + 14x^{4}y^{2})}{(9x^{3}y^{3} + 20x^{4}y^{2} + 13xy^{5} + 10x^{5}y - 4x^{6} - 5y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{352}{43}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{16}{301}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{32}{43}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{64}{43}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{32}{43}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(18(0)^{3}y^{3} + 14(0)^{4}y^{2})}{(9(0)^{3}y^{3} + 20(0)^{4}y^{2} + 13(0)y^{5} + 10(0)^{5}y - 4(0)^{6} - 5y^{6})}=\frac{0}{5y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(18x^{3}(x)^{3} + 14x^{4}(x)^{2})}{(9x^{3}(x)^{3} + 20x^{4}(x)^{2} + 13x(x)^{5} + 10x^{5}(x) - 4x^{6} - 5(x)^{6})}=-\frac{32}{43}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #281 : Limits

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1241 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}\frac{(xy^{2} + 9y^{3})}{(7xy^{2} + 5x^{3} - y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{50}{11}\\&y=x:-153\end{align*}$

$\begin{align*}&\text{y-axis: }-9\\&y=x:\frac{10}{11}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{9}{13}\\&y=x:\frac{10}{143}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{40}{11}\\&y=x:180\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-9\\&y=x:\frac{10}{11}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{((0)y^{2} + 9y^{3})}{(7(0)y^{2} + 5(0)^{3} - y^{3})}=\frac{-9y^{3}}{y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{-9y^{3}}{y^{3}}=-9\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(x(x)^{2} + 9(x)^{3})}{(7x(x)^{2} + 5x^{3} - (x)^{3})}=\frac{10}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #282 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}\frac{(6y^{4})}{(18x^{4} - 9x^{3}y + 13y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{6}{13}\\&y=x:\frac{3}{11}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{1}{39}\\&y=x:\frac{3}{121}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{15}{11}\\&y=x:\frac{48}{13}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{12}{11}\\&y=x:-\frac{42}{13}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }\frac{6}{13}\\&y=x:\frac{3}{11}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(6y^{4})}{(18(0)^{4} - 9(0)^{3}y + 13y^{4})}=\frac{6y^{4}}{13y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{6y^{4}}{13y^{4}}=\frac{6}{13}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(6(x)^{4})}{(18x^{4} - 9x^{3}(x) + 13(x)^{4})}=\frac{3}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1241 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}\frac{(17x^{2}y^{2} + 14xy^{3} - 5y^{4})}{(8x^{3}y - 15x^{2}y^{2} + 3x^{4} + 2y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{5}{2}\\&y=x:-13\end{align*}$

$\begin{align*}&\text{y-axis: }117\\&y=x:45\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{5}{34}\\&y=x:-\frac{13}{4}\end{align*}$

$\begin{align*}&\text{y-axis: }-156\\&y=x:-\frac{95}{2}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{5}{2}\\&y=x:-13\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(17(0)^{2}y^{2} + 14(0)y^{3} - 5y^{4})}{(8(0)^{3}y - 15(0)^{2}y^{2} + 3(0)^{4} + 2y^{4})}=\frac{-5y^{4}}{2y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{-5y^{4}}{2y^{4}}=-\frac{5}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(17x^{2}(x)^{2} + 14x(x)^{3} - 5(x)^{4})}{(8x^{3}(x) - 15x^{2}(x)^{2} + 3x^{4} + 2(x)^{4})}=-13\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #281 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}\frac{(18xy^{3})}{(19x^{4} + 18y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{54}{37}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{18}{37}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{252}{37}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{6}{185}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{18}{37}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(18(0)y^{3})}{(19(0)^{4} + 18y^{4})}=\frac{0}{-18y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{-18y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(18x(x)^{3})}{(19x^{4} + 18(x)^{4})}=\frac{18}{37}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #283 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(8x^{2}y^{2} - 13x^{3}y + 5x^{4})}{(2x^{4} + 2y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{34}{9}\\&y=x:-\frac{43}{13}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{46}{11}\end{align*}$

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{2}y^{2} - 13(0)^{3}y + 5(0)^{4})}{(2(0)^{4} + 2y^{4})}=\frac{0}{2y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{2y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(8x^{2}(x)^{2} - 13x^{3}(x) + 5x^{4})}{(2x^{4} + 2(x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #1241 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}-\frac{(19x^{3}y + 5x^{4})}{(6x^{4} - 16xy^{3} + 28y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{68}{3}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{4}{21}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{16}{3}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{4}{3}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{4}{3}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(19(0)^{3}y + 5(0)^{4})}{(6(0)^{4} - 16(0)y^{3} + 28y^{4})}=\frac{0}{28y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{28y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(19x^{3}(x) + 5x^{4})}{(6x^{4} - 16x(x)^{3} + 28(x)^{4})}=-\frac{4}{3}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #281 : Partial Derivatives

Example Question #1241 : Calculus 3

Example Question #281 : Partial Derivatives

Example Question #281 : Limits

Example Question #1241 : Calculus 3

Example Question #282 : Partial Derivatives

Example Question #1241 : Calculus 3

Example Question #281 : Limits

Example Question #283 : Partial Derivatives

Example Question #1241 : Calculus 3

All Calculus 3 Resources

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