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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #341 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{117}{10}\\&y=x:\frac{8}{9}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{1}{108}\\&y=x:\frac{9}{170}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{63}{10}\\&y=x:-\frac{2}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{1}{9}\\&y=x:\frac{9}{10}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }\frac{1}{9}\\&y=x:\frac{9}{10}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(20(0)y^{3} - 3(0)^{4} + y^{4})}{(13(0)^{2}y^{2} - 2(0)^{4} + 9y^{4})}=\frac{-y^{4}}{-9y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{-y^{4}}{-9y^{4}}=\frac{1}{9}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(20x(x)^{3} - 3x^{4} + (x)^{4})}{(13x^{2}(x)^{2} - 2x^{4} + 9(x)^{4})}=\frac{9}{10}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #342 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{27}{4}\\&y=x:0\end{align*}$

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

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

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{9}{52}\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #344 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(18xy^{5} + 4x^{5}y)}{(11x^{3}y^{3} + xy^{5} - x^{6} - 4y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{11}{70}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{286}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{374}{7}\\&y=x:0\end{align*}$

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(18(0)y^{5} + 4(0)^{5}y)}{(11(0)^{3}y^{3} + (0)y^{5} - (0)^{6} - 4y^{6})}=\frac{0}{4y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(18x(x)^{5} + 4x^{5}(x))}{(11x^{3}(x)^{3} + x(x)^{5} - x^{6} - 4(x)^{6})}=-\frac{22}{7}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #343 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}\frac{(5x^{6})}{(3x^{6} - 25x^{3}y^{3} + y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{5}{21}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{5}{189}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{50}{21}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{5}{21}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(5(0)^{6})}{(3(0)^{6} - 25(0)^{3}y^{3} + y^{6})}=\frac{0}{y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(5x^{6})}{(3x^{6} - 25x^{3}(x)^{3} + (x)^{6})}=-\frac{5}{21}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #344 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}\frac{(16x^{3}y^{2} - 17x^{2}y^{3} + 8y^{5})}{(25x^{3}y^{2} + 19x^{4}y + 3x^{5} + 2y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{12}{7}\\&y=x:72\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{4}{7}\\&y=x:\frac{1}{63}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{12}{7}\\&y=x:-44\end{align*}$

$\begin{align*}&\text{y-axis: }4\\&y=x:\frac{1}{7}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }4\\&y=x:\frac{1}{7}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(16(0)^{3}y^{2} - 17(0)^{2}y^{3} + 8y^{5})}{(25(0)^{3}y^{2} + 19(0)^{4}y + 3(0)^{5} + 2y^{5})}=\frac{8y^{5}}{2y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{8y^{5}}{2y^{5}}=4\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(16x^{3}(x)^{2} - 17x^{2}(x)^{3} + 8(x)^{5})}{(25x^{3}(x)^{2} + 19x^{4}(x) + 3x^{5} + 2(x)^{5})}=\frac{1}{7}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #345 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{63}{2}\\&y=x:-19\end{align*}$

$\begin{align*}&\text{y-axis: }-56\\&y=x:15\end{align*}$

$\begin{align*}&\text{y-axis: }1\\&y=x:-\frac{7}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{1}{15}\\&y=x:-\frac{7}{12}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }1\\&y=x:-\frac{7}{2}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{3} - 2(0)y^{2} + 5y^{3})}{(7(0)y^{2} - 5y^{3})}=\frac{-5y^{3}}{-5y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{-5y^{3}}{-5y^{3}}=1\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(4x^{3} - 2x(x)^{2} + 5(x)^{3})}{(7x(x)^{2} - 5(x)^{3})}=-\frac{7}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #346 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(16xy^{4} - 20x^{2}y^{3} + 4x^{4}y)}{(4x^{5} + 5y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{23}{14}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{13}{5}\\&y=x:-\frac{37}{16}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{37}{18}\\&y=x:\frac{47}{10}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(16(0)y^{4} - 20(0)^{2}y^{3} + 4(0)^{4}y)}{(4(0)^{5} + 5y^{5})}=\frac{0}{5y^{5}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{5y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(16x(x)^{4} - 20x^{2}(x)^{3} + 4x^{4}(x))}{(4x^{5} + 5(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #347 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(17x^{2}y^{2} + 17xy^{3} + 17x^{3}y)}{(5x^{4} - 6xy^{3} + 17y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{459}{16}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{153}{4}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{51}{224}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{51}{16}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{51}{16}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(17(0)^{2}y^{2} + 17(0)y^{3} + 17(0)^{3}y)}{(5(0)^{4} - 6(0)y^{3} + 17y^{4})}=\frac{0}{17y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{17y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(17x^{2}(x)^{2} + 17x(x)^{3} + 17x^{3}(x))}{(5x^{4} - 6x(x)^{3} + 17(x)^{4})}=-\frac{51}{16}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #348 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{29}{4}\\&y=x:-\frac{31}{15}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{41}{14}\\&y=x:\frac{5}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{27}{19}\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #351 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{10}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{51}{2}\\&y=x:0\end{align*}$

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

$\begin{align*}&\text{y-axis: }0\\&y=x:-24\end{align*}$

Correct answer:

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

Explanation:

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #341 : Partial Derivatives

Example Question #342 : Partial Derivatives

Example Question #344 : Limits

Example Question #343 : Partial Derivatives

Example Question #344 : Partial Derivatives

Example Question #345 : Partial Derivatives

Example Question #346 : Partial Derivatives

Example Question #347 : Partial Derivatives

Example Question #348 : Partial Derivatives

Example Question #351 : Partial Derivatives

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