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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1252 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(7x^{3}y - 16xy^{3} + 9x^{4})}{(5x^{4} + 3y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{38}{3}\\&y=x:-\frac{13}{4}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{43}{16}\\&y=x:\frac{44}{7}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{25}{17}\\&y=x:-\frac{35}{17}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(7(0)^{3}y - 16(0)y^{3} + 9(0)^{4})}{(5(0)^{4} + 3y^{4})}=\frac{0}{3y^{4}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{3y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(7x^{3}(x) - 16x(x)^{3} + 9x^{4})}{(5x^{4} + 3(x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #302 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(3xy^{2})}{(12x^{2}y + 8x^{3} - 3y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{17}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{60}{17}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{36}{17}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{187}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{17}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(3(0)y^{2})}{(12(0)^{2}y + 8(0)^{3} - 3y^{3})}=\frac{0}{3y^{3}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(3x(x)^{2})}{(12x^{2}(x) + 8x^{3} - 3(x)^{3})}=-\frac{3}{17}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #303 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(2x^{4}y^{2} + 12xy^{5} - 14x^{5}y)}{(5x^{6} + 3y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{23}{2}\\&y=x:-\frac{37}{12}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{5}{3}\\&y=x:0\end{align*}$

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

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{23}{5}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)^{4}y^{2} + 12(0)y^{5} - 14(0)^{5}y)}{(5(0)^{6} + 3y^{6})}=\frac{0}{3y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(2x^{4}(x)^{2} + 12x(x)^{5} - 14x^{5}(x))}{(5x^{6} + 3(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #301 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{54}{5}\\&y=x:0\end{align*}$

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

$\begin{align*}&\text{y-axis: }\frac{84}{5}\\&y=x:0\end{align*}$

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

Correct answer:

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

Explanation:

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

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Example Question #1261 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}\frac{(11xy^{5} + 14x^{6})}{(20x^{5}y - 5x^{6} + 5y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

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

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

$\begin{align*}&\text{y-axis: }-\frac{45}{4}\\&y=x:0\end{align*}$

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(11(0)y^{5} + 14(0)^{6})}{(20(0)^{5}y - 5(0)^{6} + 5y^{6})}=\frac{0}{-5y^{6}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{-5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(11x(x)^{5} + 14x^{6})}{(20x^{5}(x) - 5x^{6} + 5(x)^{6})}=\frac{5}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #306 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(16xy^{2} - 15x^{3})}{(2x^{3} - 15x^{2}y + 11y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{1}{16}\end{align*}$

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

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

$\begin{align*}&\text{y-axis: }\frac{11}{2}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(16(0)y^{2} - 15(0)^{3})}{(2(0)^{3} - 15(0)^{2}y + 11y^{3})}=\frac{0}{11y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{11y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(16x(x)^{2} - 15x^{3})}{(2x^{3} - 15x^{2}(x) + 11(x)^{3})}=\frac{1}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #301 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{57}\end{align*}$

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

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{17}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{1}{3}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{1}{3}\\&y=x:0\end{align*}$

Explanation:

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

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Example Question #1262 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(8x^{4}y + 6x^{5})}{(10x^{3}y^{2} + 19x^{4}y + 5x^{5} + y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{26}{5}\\&y=x:0\end{align*}$

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

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{4}y + 6(0)^{5})}{(10(0)^{3}y^{2} + 19(0)^{4}y + 5(0)^{5} + y^{5})}=\frac{0}{y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(8x^{4}(x) + 6x^{5})}{(10x^{3}(x)^{2} + 19x^{4}(x) + 5x^{5} + (x)^{5})}=-\frac{2}{5}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #309 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(3x^{2}y^{3} - 20x^{3}y^{2} + 4x^{5})}{(17xy^{4} + 9x^{5} - 15y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{13}{11}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{1}{11}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{52}{11}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{234}{11}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{13}{11}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(3(0)^{2}y^{3} - 20(0)^{3}y^{2} + 4(0)^{5})}{(17(0)y^{4} + 9(0)^{5} - 15y^{5})}=\frac{0}{-15y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{-15y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(3x^{2}(x)^{3} - 20x^{3}(x)^{2} + 4x^{5})}{(17x(x)^{4} + 9x^{5} - 15(x)^{5})}=\frac{13}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1263 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(14y^{4})}{(10x^{2}y^{2} + 8x^{3}y + 9x^{4} + 2y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{168}{29}\\&y=x:-28\end{align*}$

$\begin{align*}&\text{y-axis: }-7\\&y=x:-\frac{14}{29}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{42}{29}\\&y=x:133\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{7}{20}\\&y=x:-\frac{7}{87}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-7\\&y=x:-\frac{14}{29}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(14y^{4})}{(10(0)^{2}y^{2} + 8(0)^{3}y + 9(0)^{4} + 2y^{4})}=\frac{-14y^{4}}{2y^{4}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{-14y^{4}}{2y^{4}}=-7\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(14(x)^{4})}{(10x^{2}(x)^{2} + 8x^{3}(x) + 9x^{4} + 2(x)^{4})}=-\frac{14}{29}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1252 : Calculus 3

Example Question #302 : Limits

Example Question #303 : Limits

Example Question #301 : Partial Derivatives

Example Question #1261 : Calculus 3

Example Question #306 : Limits

Example Question #301 : Limits

Example Question #1262 : Calculus 3

Example Question #309 : Limits

Example Question #1263 : Calculus 3

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