Partial Differentiation - Multivariable Calculus

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

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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}$$

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

Tap to see back →

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}$$

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

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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:

Power Functions:

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

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

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

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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:

Power Functions:

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

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

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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}$$

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

Tap to see back →

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}$$

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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}$$

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

Tap to see back →

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}$$

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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}$$

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

Tap to see back →

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}$$

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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:

Power Functions:

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

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

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

Tap to see back →

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:

Power Functions:

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

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