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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #241 : Partial Derivatives

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

Possible Answers:

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

$\begin{align*}&\text{x-axis: }-\frac{16}{5}\\&y=x:-\frac{37}{19}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{32}{9}\\&y=x:-5\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{45}{19}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{3}y^{3} + 5(0)^{4}y^{2} - 9(0)^{5}y)}{((0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(4x^{3}(x)^{3} + 5x^{4}(x)^{2} - 9x^{5}(x))}{(x^{6} + 2(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #241 : Limits

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #241 : Limits

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

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{31}{17}\\&y=x:-\frac{33}{5}\end{align*}$

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

$\begin{align*}&\text{x-axis: }-\frac{19}{5}\\&y=x:\frac{47}{18}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{36}{17}\\&y=x:-\frac{11}{3}\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #242 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(11xy^{4} + 11x^{4}y + 2y^{5})}{(3x^{2}y^{3} + 18x^{3}y^{2} + 9x^{5} + 5y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{1}{30}\\&y=x:\frac{12}{245}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{432}{35}\\&y=x:-6\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{312}{35}\\&y=x:\frac{28}{5}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{2}{5}\\&y=x:\frac{24}{35}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }\frac{2}{5}\\&y=x:\frac{24}{35}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(11(0)y^{4} + 11(0)^{4}y + 2y^{5})}{(3(0)^{2}y^{3} + 18(0)^{3}y^{2} + 9(0)^{5} + 5y^{5})}=\frac{2y^{5}}{5y^{5}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{2y^{5}}{5y^{5}}=\frac{2}{5}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(11x(x)^{4} + 11x^{4}(x) + 2(x)^{5})}{(3x^{2}(x)^{3} + 18x^{3}(x)^{2} + 9x^{5} + 5(x)^{5})}=\frac{24}{35}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #245 : Limits

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

Possible Answers:

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

$\begin{align*}&\text{y-axis: }\frac{21}{5}\\&y=x:-2\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{9}{4}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{31}{2}\\&y=x:-\frac{31}{5}\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #246 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{28}{5}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{34}{5}\\&y=x:-\frac{21}{4}\end{align*}$

$\begin{align*}&\text{y-axis: }-7\\&y=x:\frac{19}{8}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)^{2}y^{4} + 14(0)y^{5} - 16(0)^{6})}{(4(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(2x^{2}(x)^{4} + 14x(x)^{5} - 16x^{6})}{(4x^{6} + 4(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #247 : Limits

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

Possible Answers:

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

$\begin{align*}&\text{y-axis: }-\frac{45}{2}\\&y=x:0\end{align*}$

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

$\begin{align*}&\text{y-axis: }0\\&y=x:50\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #242 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(15x^{2}y^{4} + 13x^{3}y^{3} + 3y^{6})}{(20x^{3}y^{3} - 9x^{2}y^{4} + 20x^{5}y + 2x^{6} - 4y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{217}{29}\\&y=x:-\frac{21}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{434}{29}\\&y=x:\frac{9}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{3}{80}\\&y=x:-\frac{31}{348}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{3}{4}\\&y=x:-\frac{31}{29}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }\frac{3}{4}\\&y=x:-\frac{31}{29}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(15(0)^{2}y^{4} + 13(0)^{3}y^{3} + 3y^{6})}{(20(0)^{3}y^{3} - 9(0)^{2}y^{4} + 20(0)^{5}y + 2(0)^{6} - 4y^{6})}=\frac{-3y^{6}}{-4y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{-3y^{6}}{-4y^{6}}=\frac{3}{4}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(15x^{2}(x)^{4} + 13x^{3}(x)^{3} + 3(x)^{6})}{(20x^{3}(x)^{3} - 9x^{2}(x)^{4} + 20x^{5}(x) + 2x^{6} - 4(x)^{6})}=-\frac{31}{29}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #243 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{1}{5}\\&y=x:\frac{1}{14}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-7\\&y=x:\frac{12}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{2}{5}\\&y=x:\frac{1}{2}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{2}{5}\\&y=x:\frac{1}{2}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(10(0)y^{2} + 2y^{3})}{(19(0)^{2}y + 10(0)^{3} - 5y^{3})}=\frac{-2y^{3}}{5y^{3}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{-2y^{3}}{5y^{3}}=-\frac{2}{5}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(10x(x)^{2} + 2(x)^{3})}{(19x^{2}(x) + 10x^{3} - 5(x)^{3})}=\frac{1}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #250 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{34}{3}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{31}{13}\\&y=x:-\frac{26}{15}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{47}{9}\\&y=x:-\frac{3}{2}\end{align*}$

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

Correct answer:

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

Explanation:

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

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #241 : Partial Derivatives

Example Question #241 : Limits

Example Question #241 : Limits

Example Question #242 : Limits

Example Question #245 : Limits

Example Question #246 : Limits

Example Question #247 : Limits

Example Question #242 : Limits

Example Question #243 : Limits

Example Question #250 : Limits

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