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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 #192 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(6x^{3}y)}{(5x^{2}y^{2} - 17xy^{3} + 8x^{3}y + x^{4} - y^{4})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(6(0)^{3}y)}{(5(0)^{2}y^{2} - 17(0)y^{3} + 8(0)^{3}y + (0)^{4} - y^{4})}=\frac{0}{y^{4}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(6x^{3}(x))}{(5x^{2}(x)^{2} - 17x(x)^{3} + 8x^{3}(x) + x^{4} - (x)^{4})}=-\frac{3}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #193 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{23}{6}\\&y=x:-\frac{27}{7}\end{align*}$

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

$\begin{align*}&\text{x-axis: }\frac{22}{19}\\&y=x:-\frac{27}{8}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{24}{7}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #1151 : Calculus 3

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

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{28}{11}\\&y=x:\frac{25}{7}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{21}{4}\\&y=x:-\frac{42}{17}\end{align*}$

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

$\begin{align*}&\text{x-axis: }-19\\&y=x:-3\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(18(0)^{4}y^{2} - 18(0)^{5}y)}{(2(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(18x^{4}(x)^{2} - 18x^{5}(x))}{(2x^{6} + 4(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #195 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}-\frac{(8x^{2}y^{2} - 7x^{3}y)}{(12xy^{3} + 36x^{4} + 4y^{4})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{1}{52}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{52}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{1}{52}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{1}{52}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{52}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{2}y^{2} - 7(0)^{3}y)}{(12(0)y^{3} + 36(0)^{4} + 4y^{4})}=\frac{0}{4y^{4}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{4y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(8x^{2}(x)^{2} - 7x^{3}(x))}{(12x(x)^{3} + 36x^{4} + 4(x)^{4})}=-\frac{1}{52}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1151 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}\frac{(11x^{4}y^{2} - 17x^{6})}{(2x^{4}y^{2} - 12x^{3}y^{3} - 19x^{5}y + 17x^{6} + y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{6}{11}\\&y=x:0\end{align*}$

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

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{6}{11}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{6}{11}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{6}{11}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(11(0)^{4}y^{2} - 17(0)^{6})}{(2(0)^{4}y^{2} - 12(0)^{3}y^{3} - 19(0)^{5}y + 17(0)^{6} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(11x^{4}(x)^{2} - 17x^{6})}{(2x^{4}(x)^{2} - 12x^{3}(x)^{3} - 19x^{5}(x) + 17x^{6} + (x)^{6})}=\frac{6}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #197 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{16}{3}\\&y=x:-\frac{14}{9}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{45}{8}\\&y=x:-\frac{5}{2}\end{align*}$

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

$\begin{align*}&\text{x-axis: }-\frac{15}{4}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(9(0)^{4}y^{2} + 9(0)y^{5} - 18(0)^{6})}{(3(0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(9x^{4}(x)^{2} + 9x(x)^{5} - 18x^{6})}{(3x^{6} + 2(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #1151 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(6x^{2}y^{3} + 3x^{3}y^{2})}{(17x^{3}y^{2} - 8x^{2}y^{3} + 2xy^{4} - 20x^{4}y + x^{5} + 4y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

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

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{1}{8}\end{align*}$

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

$\begin{align*}&\text{x-axis: }-\frac{117}{4}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(6(0)^{2}y^{3} + 3(0)^{3}y^{2})}{(17(0)^{3}y^{2} - 8(0)^{2}y^{3} + 2(0)y^{4} - 20(0)^{4}y + (0)^{5} + 4y^{5})}=\frac{0}{4y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(6x^{2}(x)^{3} + 3x^{3}(x)^{2})}{(17x^{3}(x)^{2} - 8x^{2}(x)^{3} + 2x(x)^{4} - 20x^{4}(x) + x^{5} + 4(x)^{5})}=\frac{9}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #191 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}-\frac{(10x^{3}y^{3} + 2x^{6})}{(6x^{2}y^{4} + 11x^{4}y^{2} + 26x^{5}y - 3x^{6} - 3y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{240}{37}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{4}{185}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{156}{37}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{12}{37}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{12}{37}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(10(0)^{3}y^{3} + 2(0)^{6})}{(6(0)^{2}y^{4} + 11(0)^{4}y^{2} + 26(0)^{5}y - 3(0)^{6} - 3y^{6})}=\frac{0}{3y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(10x^{3}(x)^{3} + 2x^{6})}{(6x^{2}(x)^{4} + 11x^{4}(x)^{2} + 26x^{5}(x) - 3x^{6} - 3(x)^{6})}=-\frac{12}{37}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #200 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(x^{4}y)}{(15x^{4}y - 14x^{3}y^{2} - 16x^{2}y^{3} + 3x^{5} + y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

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

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{1}{198}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{17}{11}\\&y=x:0\end{align*}$

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)^{4}y)}{(15(0)^{4}y - 14(0)^{3}y^{2} - 16(0)^{2}y^{3} + 3(0)^{5} + y^{5})}=\frac{0}{y^{5}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(x^{4}(x))}{(15x^{4}(x) - 14x^{3}(x)^{2} - 16x^{2}(x)^{3} + 3x^{5} + (x)^{5})}=\frac{1}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #201 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}\frac{(4xy^{2} - 17x^{2}y + 12x^{3})}{(12xy^{2} + 18x^{2}y - x^{3} + 3y^{3})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{7}{32}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{3}{32}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{352}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{32}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{32}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(4(0)y^{2} - 17(0)^{2}y + 12(0)^{3})}{(12(0)y^{2} + 18(0)^{2}y - (0)^{3} + 3y^{3})}=\frac{0}{3y^{3}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(4x(x)^{2} - 17x^{2}(x) + 12x^{3})}{(12x(x)^{2} + 18x^{2}(x) - x^{3} + 3(x)^{3})}=-\frac{1}{32}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #192 : Partial Derivatives

Example Question #193 : Partial Derivatives

Example Question #1151 : Calculus 3

Example Question #195 : Partial Derivatives

Example Question #1151 : Calculus 3

Example Question #197 : Partial Derivatives

Example Question #1151 : Calculus 3

Example Question #191 : Partial Derivatives

Example Question #200 : Partial Derivatives

Example Question #201 : Partial Derivatives

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