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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 #1063 : Partial Derivatives

Find the derivative of the function with respect to by taking the natural logarithm of both sides and then differentiating both sides with respect to .

$z=\frac{xy+2xy}{4y\sqrt{y^2+2x}}$

Possible Answers:

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x^2}{x^3y+xy} +\frac{5}{5y+x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x+2x}{xy+2xy} +\frac{4}{x}+\frac{4x+1}{2x^2+x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{3y}{xy+2xy} +\frac{4y+1}{2y^2+y} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left(\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x}{x^3y+x} +\frac{5y}{5y^2+x} \right )$

Correct answer:

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left(\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right )$

Explanation:

$z=f(x,y)=\frac{xy+2xy}{4y\sqrt{y^2+2x}}$

Take the natural logarithm of both sides and expand the right-side using the properties of logarithms.

$\ln(z)=\ln\left(\frac{xy+2xy}{4y\sqrt{y^2+2x}}\right)$

Apply the rule for the logarithm of a quotient:

$\ln(z)=\ln\left(xy+2xy\right)-\ln\left(4y\sqrt{y^2+2x} \right)$

Apply the rule for the logarithm of a product:

$\ln(z)=\ln\left(xy+2xy\right)-\left[\ln(4y)+\ln\left(\sqrt{y^2+2x}\right) \right]$

Apply the rule for the logarithm of a quantity raised to a power:

$\ln(z)=\ln\left(xy+2xy\right)-\ln(4y)-\frac{1}{2}\ln\left(y^2+2x\right)$

Now differentiate both sides implicitly, remembering that is a multi-variable function of and .

$\frac{1}{z}\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}\left[\ln\left(xy+2xy\right)-\ln(4y)-\frac{1}{2}\ln\left(y^2+2x\right)\right]$

Proceeding with the differentiation on the right-side with respect to .

$\frac{1}{z}\frac{\partial z}{\partial y}=\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x}$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left[\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right ]$

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Example Question #1064 : Partial Derivatives

Find $\frac{\partial f}{dy}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dy}=x+\frac{z}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+ye^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

$\frac{\partial f}{dy}=\frac{x}{\ln(yz)}$

$\frac{\partial f}{dy}=xz^{10}$

$\frac{\partial f}{dy}=z^{10}+\frac{1}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+e^{xy}$

Correct answer:

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

Explanation:

In order to find $\frac{\partial f}{dy}$ , we need to take the derivative of in respect to , and treat , and as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dy}=xz^{10}+x(\ln(yz))^{-1}-xy(\ln(yz))^{-2}\cdot\frac{z}{yz}+0+xe^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

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Example Question #1065 : Partial Derivatives

True or False

$\frac{\partial}{\partial y}\Big(x^x+wz\ln(uvxw)\Big)=0$

Possible Answers:

False

True

Correct answer:

True

Explanation:

True:

Since there are no y's in the equation, the derivative of a constant is .

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Example Question #1061 : Partial Derivatives

Find $\frac{\partial f}{dy}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dy}=x+\frac{z}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+ye^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

$\frac{\partial f}{dy}=z^{10}+\frac{1}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+e^{xy}$

$\frac{\partial f}{dy}=xz^{10}$

$\frac{\partial f}{dy}=\frac{x}{\ln(yz)}$

Correct answer:

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

Explanation:

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

$f'(x)=nx^{n-1}$

$\frac{\partial f}{dy}=xz^{10}+x(\ln(yz))^{-1}-xy(\ln(yz))^{-2}\cdot\frac{z}{yz}+0+xe^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

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Example Question #1062 : Partial Derivatives

True or False

$\frac{\partial}{\partial y}\Big(x^x+wz\ln(uvxw)\Big)=0$

Possible Answers:

True

False

Correct answer:

True

Explanation:

True:

Since there are no y's in the equation, the derivative of a constant is .

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Example Question #1 : Directional Derivatives

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=x^2y+xz^{10}+xy^2z^5$ in the direction of $\vec{v}=(1,2,3)$ .

Possible Answers:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((2xy+z^{10}+y^2z^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= 2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4)$

Correct answer:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

Explanation:

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\neq1$

$\vec{u}=\frac{1}{\sqrt{14}}\vec{v}=\Big(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+0+2xyz^5)+\frac{3}{\sqrt{14}}(0+10xz^9+5xy^2z^4)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+2xyz^5)+\frac{3}{\sqrt{14}}(10xz^9+5xy^2z^4)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

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Example Question #2 : Directional Derivatives

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=z^2y\ln(x)+xy^2z^3$ in the direction of $\vec{v}=(0,1,1)$ .

Possible Answers:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ (2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ (z^2\ln(x)+2xyz^3)$

Correct answer:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

Explanation:

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{0^2+1^2+1^2}=\sqrt{0+1+1}=\sqrt{2}\neq1$

$\vec{u}=\frac{1}{\sqrt{2}}\vec{v}=\Big(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ 0(\frac{yz^2}{x}+y^2z^3)+\frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

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Example Question #3 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-5,1,4)\text{ where }f(x,y,z)=y^3cos{(x^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-2,20,-14).\end{align*}$

Possible Answers:

23.13

-1.55

18.92

-66.53

Correct answer:

-1.55

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-5,1,4)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(20)^2+(-14)^2}=24.49\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{24.49}=-0.08\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{24.49}=0.82\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{24.49}=-0.57\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-5,1,4)=(-0.08)(-2xy^3sin{(x^2)}sin{(z)})+(0.82)(3y^2cos{(x^2)}sin{(z)})+(-0.57)(y^3cos{(x^2)}cos{(z)})=-1.55\end{align*}$

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Example Question #1 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(4,2,-1)\text{ where }f(x,y,z)=x^3sin{(y^2)}sin{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(-16,4,-11).\end{align*}$

Possible Answers:

2.5

47.3

-79.36

6.12

Correct answer:

47.3

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(4,2,-1)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(4)^2+(-11)^2}=19.82\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{19.82}=-0.81\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{19.82}=0.2\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{19.82}=-0.55\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(4,2,-1)=(-0.81)(3x^2sin{(y^2)}sin{(z^3)})+(0.2)(2x^3ycos{(y^2)}sin{(z^3)})+(-0.55)(3x^3z^2cos{(z^3)}sin{(y^2)})=47.3\end{align*}$

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Example Question #2 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-1,1,-3)\text{ where }f(x,y,z)=x^4z^4cos{(y^2)}\\&\text{In the direction of }\overrightarrow{v}=(13,20,9).\end{align*}$

Possible Answers:

2.51

1.37

-216.8

138.98

Correct answer:

-216.8

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-1,1,-3)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(20)^2+(9)^2}=25.5\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{25.5}=0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{25.5}=0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{25.5}=0.35\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-1,1,-3)=(0.51)(4x^3z^4cos{(y^2)})+(0.78)(-2x^4yz^4sin{(y^2)})+(0.35)(4x^4z^3cos{(y^2)})=-216.8\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1063 : Partial Derivatives

Example Question #1064 : Partial Derivatives

Example Question #1065 : Partial Derivatives

Example Question #1061 : Partial Derivatives

Example Question #1062 : Partial Derivatives

Example Question #1 : Directional Derivatives

Example Question #2 : Directional Derivatives

Example Question #3 : Directional Derivatives

Example Question #1 : Directional Derivatives

Example Question #2 : Directional Derivatives

All Calculus 3 Resources

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