Multivariable Calculus - Multivariable Calculus

Card 0 of 88

Question

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

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

Answer

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:

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

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

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

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Question

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

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

Answer

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:

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

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

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

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Question

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

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

Answer

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:

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

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

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

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Question

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Answer

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r^2\sin(\theta) \ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} zr^2\sin(\theta) \Big|_{0}^{r\cos(\theta)+1} \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} (r\cos(\theta)+1)r^2\sin(\theta)-0 \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} r^3\sin(\theta)\cos(\theta)+r^2\sin(\theta) \ dr \ d\theta$

$=\int_{0}^{2\pi} \frac{1}{4}r^4\sin(\theta)\cos(\theta)+\frac{1}{3}r^3\sin(\theta) \Big|_{1}^{3}\ d\theta$

$=\int_{0}^{2\pi} (\frac{1}{4}(3)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(3)^3\sin(\theta))-(\frac{1}{4}(1)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(1)^3\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} (\frac{81}{4}\sin(\theta)\cos(\theta)+9\sin(\theta))-(\frac{1}{4}\sin(\theta)\cos(\theta)+\frac{1}{3}\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} 20\sin(\theta)\cos(\theta)+\frac{26}{3}\sin(\theta) \ d\theta$

$= -\frac{1}{2}\cdot20\cos^2(\theta)-\frac{26}{3}\cos(\theta)\Big|_{0}^{2\pi}$

$= -\frac{1}{2}\cdot20\cos^2(2\pi)-\frac{26}{3}\cos(2\pi)-(-\frac{1}{2}\cdot20\cos^2(0)-\frac{26}{3}\cos(0))$

$= -\frac{1}{2}\cdot20(1)-\frac{26}{3}(1)-(-\frac{1}{2}\cdot20(1)-\frac{26}{3}(1))$

$= -10-\frac{26}{3}-(-10-\frac{26}{3})=0$

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Question

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Answer

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r^2\sin(\theta) \ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} zr^2\sin(\theta) \Big|_{0}^{r\cos(\theta)+1} \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} (r\cos(\theta)+1)r^2\sin(\theta)-0 \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} r^3\sin(\theta)\cos(\theta)+r^2\sin(\theta) \ dr \ d\theta$

$=\int_{0}^{2\pi} \frac{1}{4}r^4\sin(\theta)\cos(\theta)+\frac{1}{3}r^3\sin(\theta) \Big|_{1}^{3}\ d\theta$

$=\int_{0}^{2\pi} (\frac{1}{4}(3)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(3)^3\sin(\theta))-(\frac{1}{4}(1)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(1)^3\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} (\frac{81}{4}\sin(\theta)\cos(\theta)+9\sin(\theta))-(\frac{1}{4}\sin(\theta)\cos(\theta)+\frac{1}{3}\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} 20\sin(\theta)\cos(\theta)+\frac{26}{3}\sin(\theta) \ d\theta$

$= -\frac{1}{2}\cdot20\cos^2(\theta)-\frac{26}{3}\cos(\theta)\Big|_{0}^{2\pi}$

$= -\frac{1}{2}\cdot20\cos^2(2\pi)-\frac{26}{3}\cos(2\pi)-(-\frac{1}{2}\cdot20\cos^2(0)-\frac{26}{3}\cos(0))$

$= -\frac{1}{2}\cdot20(1)-\frac{26}{3}(1)-(-\frac{1}{2}\cdot20(1)-\frac{26}{3}(1))$

$= -10-\frac{26}{3}-(-10-\frac{26}{3})=0$

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Question

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

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

Answer

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:

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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:

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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:

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

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

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

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Question

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

Answer

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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Question

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

Answer

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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Question

Write down the equation of the line in vector form that passes through the points , and .

Answer

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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Question

Write down the equation of the line in vector form that passes through the points , and .

Answer

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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Question

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

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

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

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

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

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

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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Question

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

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

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

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

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

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

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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Question

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

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

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

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

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

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

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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Question

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

Answer

First, we need to find the partial derivatives in respect to , and , and plug in .

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

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

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

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

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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Question

Let , and .

Find $u\times v$ .

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , 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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Question

Let , and .

Find $u\times v$ .

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , 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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Question

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

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

Answer

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:

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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:

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

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

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

Compare your answer with the correct one above

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