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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 #181 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-39x^{3}}{20sin(7\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$-\frac{39}{(980\cdot \pi ^{2})^{(\frac{3}{2})}}$

$\infty$

$\frac{39}{(980\cdot \pi ^{2})^{(\frac{3}{2})}}$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-39x^{3}}{20sin(7\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{-117x^{2}}{280\cdot \pi cos(7\cdot \pi x)sin(7\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{-234x}{1960\cdot \pi ^{2}cos(7\cdot \pi x)^{2} - 1960\cdot \pi ^{2}sin(7\cdot \pi x)^{2}}\rightarrow\frac{0}{1960\cdot \pi ^{2}}=0\\&lim_{x\rightarrow0-}\frac{-39x^{3}}{20sin(7\cdot \pi x)^{2}}=0\end{align*}$

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Example Question #1142 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{x^{2}}{sin(13\cdot \pi x)^{2}}\end{align*}$

Possible Answers:

$\frac{1}{(169\cdot \pi ^{2})}$

$\infty$

$-\infty$

$-\frac{1}{(169\cdot \pi ^{2})}$

Correct answer:

$\frac{1}{(169\cdot \pi ^{2})}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{x^{2}}{sin(13\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\\&\frac{2x}{26\cdot \pi cos(13\cdot \pi x)sin(13\cdot \pi x)}\rightarrow\frac{0}{0}\\&\frac{2}{338\cdot \pi ^{2}cos(13\cdot \pi x)^{2} - 338\cdot \pi ^{2}sin(13\cdot \pi x)^{2}}\rightarrow\frac{2}{338\cdot \pi ^{2}}\\&lim_{x\rightarrow0-}\frac{x^{2}}{sin(13\cdot \pi x)^{2}}=\frac{1}{(169\cdot \pi ^{2})}\end{align*}$

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Example Question #1143 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{-5x}{8sin(5\cdot \pi x)^{3}}\end{align*}$

Possible Answers:

$\frac{5}{(10\cdot \pi )}$

$\infty$

$-\infty$

$-\frac{5}{(10\cdot \pi )}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{-5x}{8sin(5\cdot \pi x)^{3}}\rightarrow\frac{0}{0}\\&\frac{-5}{120\cdot \pi cos(5\cdot \pi x)sin(5\cdot \pi x)^{2}}\rightarrow\frac{-5}{0}=-\infty\\&lim_{x\rightarrow0-}\frac{-5x}{8sin(5\cdot \pi x)^{3}}=-\infty\end{align*}$

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Example Question #181 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}\frac{9sin(6x)^{2}}{7x^{2}}\end{align*}$

Possible Answers:

$-\frac{324}{7}$

$\frac{324}{7}$

$\infty$

$-\infty$

Correct answer:

$\frac{324}{7}$

Explanation:

$\begin{align*}&\text{Examining this problem, we find that both}\\&\text{numerator and denominator have zero values at the limit specified.}\\&\text{Noting this, L'Hopital's method may be of use:}\\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\\&\frac{9sin(6x)^{2}}{7x^{2}}\rightarrow\frac{0}{0}\\&\frac{108cos(6x)sin(6x)}{14x}\rightarrow\frac{0}{0}\\&\frac{648cos(6x)^{2} - 648sin(6x)^{2}}{14}\rightarrow\frac{648}{14}\\&lim_{x\rightarrow0-}\frac{9sin(6x)^{2}}{7x^{2}}=\frac{324}{7}\end{align*}$

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Example Question #182 : Partial Derivatives

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(4x^{3}y^{3} + xy^{5} - 7x^{5}y)}{(14x^{6} - 27x^{3}y^{3} + y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

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

$\begin{align*}&\text{x-axis: }-\frac{1}{6}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{6}\end{align*}$

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

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{1}{6}\end{align*}$

Explanation:

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Example Question #1141 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}\frac{(7xy^{5} - 14x^{5}y)}{(8xy^{5} + 5x^{5}y + 16x^{6} - 3y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{7}{26}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{7}{26}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{7}{26}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{7}{26}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{7}{26}\end{align*}$

Explanation:

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Example Question #181 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{13}{21}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{13}{21}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{13}{21}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{13}{21}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{13}{21}\end{align*}$

Explanation:

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Example Question #1147 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #182 : Limits

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

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&x=y:-\frac{33}{25}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{33}{25}\\&x=y:0\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{33}{25}\\&x=y:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&x=y:\frac{33}{25}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&x=y:\frac{33}{25}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(20(0)^{3}y^{3} + 13(0)^{4}y^{2})}{(15(0)^{2}y^{4} + (0)y^{5} + 5(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(20x^{3}(x)^{3} + 13x^{4}(x)^{2})}{(15x^{2}(x)^{4} + x(x)^{5} + 5x^{6} + 4(x)^{6})}=\frac{33}{25}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$ $\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(20(0)^{3}y^{3} + 13(0)^{4}y^{2})}{(15(0)^{2}y^{4} + (0)y^{5} + 5(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }x=y:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(20x^{3}(x)^{3} + 13x^{4}(x)^{2})}{(15x^{2}(x)^{4} + x(x)^{5} + 5x^{6} + 4(x)^{6})}=\frac{33}{25}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #191 : Partial Derivatives

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

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{43}{8}\\&y=x:\frac{5}{2}\end{align*}$

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

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{37}{19}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{37}{7}\\&y=x:-\frac{39}{7}\end{align*}$

Correct answer:

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

Explanation:

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #181 : Partial Derivatives

Example Question #1142 : Calculus 3

Example Question #1143 : Calculus 3

Example Question #181 : Partial Derivatives

Example Question #182 : Partial Derivatives

Example Question #1141 : Calculus 3

Example Question #181 : Partial Derivatives

Example Question #1147 : Calculus 3

Example Question #182 : Limits

Example Question #191 : Partial Derivatives

All Calculus 3 Resources

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