In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given, the vector form is  .

So for $\displaystyle -6i+j+k$, we can derive the vector form $\displaystyle \left \langle -6,1,1\right \rangle$.

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given, the vector form is  .

So for $\displaystyle 3i-k$, we can derive the vector form $\displaystyle \left \langle 3,0,-1\right \rangle$.

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point and , the distance is the vector .

Subbing in our original points $\displaystyle (4,2,-2)$ and $\displaystyle (7,1,0)$,  we get:

$\displaystyle v=\left \langle 7-4,1-2,0-(-2)\right \rangle$

$\displaystyle v=\left \langle 3,-1,2\right \rangle$

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point and , the distance is the vector .

Subbing in our original points $\displaystyle (2,7,-6)$ and $\displaystyle (1,2,3)$,  we get:

 $\displaystyle v=\left \langle 1-2,2-7,3-(-6)\right \rangle$

$\displaystyle v=\left \langle -1,-5,9\right \rangle$

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given, the vector form is .

So for $\displaystyle -i+2j+10k$, we can derive the vector form $\displaystyle \left \langle -1,2,10\right \rangle$.

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and  coefficients.

That is, given , the vector form is .

So for $\displaystyle i+k$, we can derive the vector form $\displaystyle \left \langle 1,0,1\right \rangle$.

Given $\displaystyle j+8k$, we need to map the $\displaystyle i$, $\displaystyle j$, and $\displaystyle k$ coefficients back to their corresponding $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$-coordinates.

Thus the vector form of $\displaystyle j+8k$ is $\displaystyle \left \langle 0,1,8\right \rangle$.

In order to express $\displaystyle 4i-9j$ in vector form, we must use the coefficients of $\displaystyle i, j,$and $\displaystyle k$ to represent the $\displaystyle x$-, $\displaystyle y$-, and $\displaystyle z$-coordinates of the vector.

Therefore, its vector form is 

$\displaystyle \left \langle 4,-9,0\right \rangle$.

Arclength is given by the formula:

$\displaystyle Arclength = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}}\right)^2 dx$

We should find dy/dx first, which we find to be:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(ln(cos(x)) = \frac{1}{cos(x)}\cdot (-sin(x)) = -tan(x)$

Now let's proceed with the integral:

$\displaystyle \int_{0}^{\frac{\pi}{3}} \sqrt{1 + (-tan x)^2}dx = \int_{0}^{\frac{\pi}{3}} \sqrt{(sec^2x)}dx = \int_{0}^{\frac{\pi}{3}} sec(x)dx$

(here we apply the integration described in the hint) 

$\displaystyle = \left[ln \left | tan(x) + sec(x)\right |\bigg]_{0}^{\frac{\pi}{3}} = ln (\sqrt{3} + {2})$

 which is obtained by evaluating at both boundaries.

When you plot the graph of  $\displaystyle r = \sqrt{cos(\theta )}$ , the bounds are between $\displaystyle \theta = -\frac{\pi }{2}$  and $\displaystyle \theta = \frac{\pi }{2}$.

Use the area formula  for polar equations:

$\displaystyle Area = \int_{\theta_{1} }^{\theta_{2}} \frac{1}{2}r^2 d\theta$ 

And so we find the area to be:

$\displaystyle \frac{1}{2}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} (\sqrt{cos( \theta)})^2d\theta = \frac{1}{2}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} cos (\theta)d\theta = \frac{1}{2}[sin \theta ]_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$\displaystyle = \frac{1}{2}(1+1) = 1$