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Multivariable Calculus : Multivariable Calculus

Study concepts, example questions & explanations for Multivariable Calculus

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14 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Matrices & Vectors

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ at z=(0,5) .

Possible Answers:

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Correct answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Explanation:

First, we need to find the partial derivatives in respect to , and , and plug in z=(1,5) .

$f(x,y)=\ln(4x^3+10y^2)$ , $f(0,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ , $f_x(0,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ , $f_y(0,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

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Example Question #1 : Multivariable Calculus

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ at z=(1,5) .

Possible Answers:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Correct answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Explanation:

First, we need to find the partial derivatives in respect to , and , and plug in z=(1,5) .

$f(x,y)=\ln(4x^3+10y^2)$ , $f(1,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ , $f_x(1,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ , $f_y(1,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

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Example Question #3 : Matrices & Vectors

Let u=(2,4,5) , and v=(-2,6,10) .

Find $u\times v$ .

Possible Answers:

$u\times v=(10,-30,20)$

$u\times v=(40,-20,12)$

$u\times v=(0,-3,2)$

$u\times v=(-10,30,-20)$

$u\times v=(10,30,20)$

Correct answer:

$u\times v=(10,-30,20)$

Explanation:

We are trying to find the cross product between and .

Recall the formula for cross product.

If u=(u_x,u_y,u_z) , and v=(v_x,v_y,v_z) , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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Example Question #2 : Multivariable Calculus

Let u=(2,4,5) , and v=(-2,6,10) .

Find $u\times v$ .

Possible Answers:

$u\times v=(0,-3,2)$

$u\times v=(40,-20,12)$

$u\times v=(10,-30,20)$

$u\times v=(-10,30,-20)$

$u\times v=(10,30,20)$

Correct answer:

$u\times v=(10,-30,20)$

Explanation:

We are trying to find the cross product between and .

Recall the formula for cross product.

If u=(u_x,u_y,u_z) , and v=(v_x,v_y,v_z) , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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Example Question #1 : Vector Equations

Write down the equation of the line in vector form that passes through the points (2, 10 , -6) , and (-3, 4, 14) .

Possible Answers:

$\vec{r}=\left \langle 2+5t, 10-6t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 10+6t, -6-2t\right \rangle$

$\vec{r}=\left \langle 2-5t, 10+6t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 1+t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

Correct answer:

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

Explanation:

Remember the general equation of a line in vector form:

$\vec{r}=r_0+t\vec{v}=\left \langle x_0,y_0,z_0\right \rangle+t\left \langle a,b,c\right \rangle$ , where $\left \langle x_0,y_0,z_0\right \rangle$ is the starting point, and $\left \langle a,b,c\right \rangle$ is the difference between the start and ending points.

Lets apply this to our problem.

$\vec{r}=\left \langle 2, 10, -6\right \rangle+t\left \langle 5, 6, -20\right \rangle$

Distribute the

$\vec{r}=\left \langle 2, 10, -6\right \rangle+\left \langle 5t, 6t, -20t\right \rangle$

Now we simply do vector addition to get

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

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Example Question #3 : Multivariable Calculus

Write down the equation of the line in vector form that passes through the points (2, 10 , -6) , and (-3, 4, 14) .

Possible Answers:

$\vec{r}=\left \langle 2+5t, 10+6t, -6-2t\right \rangle$

$\vec{r}=\left \langle 2-5t, 10+6t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 1+t, -6-20t\right \rangle$

$\vec{r}=\left \langle 2+5t, 10-6t, -6-20t\right \rangle$

Correct answer:

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

Explanation:

Remember the general equation of a line in vector form:

Lets apply this to our problem.

$\vec{r}=\left \langle 2, 10, -6\right \rangle+t\left \langle 5, 6, -20\right \rangle$

Distribute the

$\vec{r}=\left \langle 2, 10, -6\right \rangle+\left \langle 5t, 6t, -20t\right \rangle$

Now we simply do vector addition to get

$\vec{r}=\left \langle 2+5t, 10+6t, -6-20t\right \rangle$

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Example Question #1 : Partial Differentiation

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

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

Possible Answers:

$\frac{\partial f}{dz}=\frac{xy}{z\ln^2(yz)}$

$\frac{\partial f}{dz}=10xyz^9$

$\frac{\partial f}{dz}=10xyz^9+10e^z$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

$\frac{\partial f}{dz}=10e^z$

Correct answer:

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

Explanation:

In order to find $\frac{\partial f}{dz}$ , 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

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

$\frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

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Example Question #2 : Partial Differentiation

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

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

Possible Answers:

$\frac{\partial f}{dx}=xyz^{10}+\frac{xy}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=xz^{10}+\frac{1}{\ln(yz)}+xe^{xy}$

$\frac{\partial f}{dx}=\frac{y}{\ln(yz)}+ye^{xy}$

Correct answer:

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Explanation:

In order to find $\frac{\partial f}{dx}$ , 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

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

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+0+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

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Example Question #4 : Multivariable Calculus

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

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

Possible Answers:

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

$\frac{\partial f}{dz}=10xyz^9$

$\frac{\partial f}{dz}=\frac{xy}{z\ln^2(yz)}$

$\frac{\partial f}{dz}=10e^z$

$\frac{\partial f}{dz}=10xyz^9+10e^z$

Correct answer:

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

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

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

$\frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

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Example Question #4 : Partial Differentiation

Determine the length of the curve $\vec{r}(t)=\left \langle t,4\sin(2t),4\cos(2t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$ .

Possible Answers:

$2\pi$

$\sqrt{65}$

$2\pi\sqrt{65}$

$\pi$

Correct answer:

$2\pi\sqrt{65}$

Explanation:

First we need to find the tangent vector, and find its magnitude.

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 8\cos(2t), -8\sin(2t)\right \rangle$

$\left \| \vec{v}(t)\right \|=\sqrt{1^2+(8\cos(2t))^2+(-8\sin(2t))^2}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64\cos^2(2t)+64\sin^2(2t)}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64(\cos^2(2t)+\sin^2(2t))}$

$\left \| \vec{v}(t)\right \|=\sqrt{1+64}=\sqrt{65}$

Now we can set up our arc length integral

$L=\int_{a}^{b}\left \| \vec{v}(t)\right \|dt$

$L=\int_{0}^{2\pi} \sqrt{65}\ dt=t\sqrt{65}\Big|_{0}^{2\pi}=2\pi\sqrt{65}$

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Multivariable Calculus : Multivariable Calculus

Study concepts, example questions & explanations for Multivariable Calculus

All Multivariable Calculus Resources

Example Questions

Example Question #1 : Matrices & Vectors

Example Question #1 : Multivariable Calculus

Example Question #3 : Matrices & Vectors

Example Question #2 : Multivariable Calculus

Example Question #1 : Vector Equations

Example Question #3 : Multivariable Calculus

Example Question #1 : Partial Differentiation

Example Question #2 : Partial Differentiation

Example Question #4 : Multivariable Calculus

Example Question #4 : Partial Differentiation

All Multivariable Calculus Resources

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