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

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

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

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

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

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

Now lets apply this to our problem

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

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

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

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

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

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

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

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

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

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

Now lets apply this to our problem

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

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

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

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

We are trying to find the cross product between $\displaystyle u$ and $\displaystyle v$.

Recall the formula for cross product.

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

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

Now apply this to our situation.

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

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

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

We are trying to find the cross product between $\displaystyle u$ and $\displaystyle v$.

Recall the formula for cross product.

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

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

Now apply this to our situation.

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

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

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

Remember the general equation of a line in vector form:

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

Lets apply this to our problem.

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

Distribute the $\displaystyle t$

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

Now we simply do vector addition to get

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

Remember the general equation of a line in vector form:

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

Lets apply this to our problem.

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

Distribute the $\displaystyle t$

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

Now we simply do vector addition to get

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

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

Natural Log:

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

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

Exponential Functions:

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

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

Power Functions:

$\displaystyle f(x)=x^n$

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

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

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

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

Natural Log:

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

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

Exponential Functions:

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

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

Power Functions:

$\displaystyle f(x)=x^n$

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

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

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

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

Natural Log:

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

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

Exponential Functions:

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

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

Power Functions:

$\displaystyle f(x)=x^n$

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

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

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

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

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

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

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

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

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

Now we can set up our arc length integral

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

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