The formula to find theta in polar form is:

\(\displaystyle \theta= tan^-^1\left(\frac{y}{x}\right)\)

Plug in the Cartesian coordinates into the equation.

\(\displaystyle \theta= tan^-^1\left(\frac{1}{-1}\right) = tan^-^1(-1) =-\frac{\pi}{4}\)

However, this angle is located in the fourth quadrant and is not in the right quadrant. Add \(\displaystyle \pi\) radians to get the correct angle since the Cartesian coordinate given is located in the second quadrant.

\(\displaystyle -\frac{\pi}{4}+\pi = \frac{3\pi}{4}\)

This is a two step problem. First step is to maximize the function to find the peak of the mountain. The next step is to use the arc length formula to find the distance he climbed.

To maximize, we'll take a derivative and set it equal to zero. 

\(\displaystyle \small \small \frac{\mathrm{d} }{\mathrm{d} x}(-x^4+2x^3-x^2-x+3)=-4x^3+6x^2-2x-1\).

Setting this equal to zero, we get \(\displaystyle \small \small x\approx-0.26\).

The derivative is positive prior to this value and negative after, so it is a max. We now must take the arc length from \(\displaystyle \small x=-1\) to \(\displaystyle \small x\approx-0.26\).

The formula for arc length is 

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

For this case, the integral becomes 

\(\displaystyle \small \small \int_{-1}^{-.26}\sqrt{1+(-4x^3+6x^2-2x-1)^2}dx\).

This will give us \(\displaystyle \small 3.33\). No units were given in the problem, so leaving the answer unitless is acceptable.

Recall the formula for length in polar coordinates is given by

\(\displaystyle L=\int_{a}^{b}\sqrt{r^2+\left(\frac{dr}{d \theta}\right)^2} d \theta\).

We were given the formula

\(\displaystyle r=cos(\theta)\).

In our case, this translates to

\(\displaystyle \\L=\int_{2}^{5}\sqrt{cos^2(\theta )+((-sin(\theta ))^2}d \theta\\ \\= \int_{2}^{5}\sqrt{1}d \theta\\ \\=3\).

Write the formulas to convert from polar to Cartesian.

\(\displaystyle x=rcos(\theta)\)

\(\displaystyle y=rsin(\theta)\)

The \(\displaystyle r\) and \(\displaystyle \theta\) values are known.  Substitute both into each equation and solve for \(\displaystyle x\) and \(\displaystyle y\).

\(\displaystyle x=(-1)cos(\frac{\pi}{4}) = -\frac{\sqrt2}{2}\)

\(\displaystyle y=(-1)sin(\frac{\pi}{4}) = -\frac{\sqrt2}{2}\)

The Cartesian coordinates are:  \(\displaystyle ( -\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})\)

Write the conversion formulas.

\(\displaystyle x^2+y^2=r^2\)

\(\displaystyle x=rcos(\theta)\)

Notice the \(\displaystyle r^2\) term.  If we multiplied by \(\displaystyle r\) on both sides of the \(\displaystyle r=-4cos(\theta)\) equation, we will get:

 \(\displaystyle r^2=-4rcos(\theta)\)

\(\displaystyle r^2=-4x\)

Substitute this back into the first equation.

\(\displaystyle x^2+y^2=-4x\)

Add \(\displaystyle 4x\) on both sides.

\(\displaystyle x^2+4x+y^2=0\)

Complete the square with the \(\displaystyle x\) terms.

\(\displaystyle (x^2+4x+4)+y^2=4\)

This would then become:

\(\displaystyle (x+2)^2+y^2 = 4\)

This is a circle centered at \(\displaystyle (-2,0)\) with a radius of 4.

The answer is:  \(\displaystyle (-2,0)\)

When converting from polar to Cartesian coordinates, we must use the formulas

\(\displaystyle x = r\cos \theta \; ;\; y=r\sin \theta\)

the values \(\displaystyle r = 3\) and \(\displaystyle \theta = \frac{\pi}{6}\) are given, so we can calculate that

\(\displaystyle x = 3\cdot \cos(\frac{\pi}{6}) = \frac{3\sqrt3}{2}\)

and \(\displaystyle y = 3\cdot \sin(\frac{\pi}{6}) = \frac{3}{2}\)

So the Cartesian coordinate form is 

\(\displaystyle \Bigg( \frac{3\sqrt3}{2} , \frac{3}{2} \Bigg)\)

\(\displaystyle y=2x\) can be immediately transformed into polar form by:

\(\displaystyle rsin(\theta)=2rcos(\theta)\). 

Dividing by \(\displaystyle r\),

\(\displaystyle sin(\theta)=2cos(\theta)\) 

Dividing both sides by \(\displaystyle cos(\theta)\)

\(\displaystyle tan(\theta)=2\)

\(\displaystyle \theta=tan^{-1}(2)\)

Solving the equations \(\displaystyle r=cos(2\Theta )\) and \(\displaystyle r=\frac{1}{2}\) yields \(\displaystyle cos(2\Theta)=\frac{1}{2}\).

Hence,

\(\displaystyle 2\Theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}.\)

Therefore, the values of \(\displaystyle \Theta\) between \(\displaystyle 0\) and \(\displaystyle 2\pi\) that satisfy both equations are:

\(\displaystyle \Theta=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}.\) 

From this, it can be deduced that there are four points of intersection between the given curves:

\(\displaystyle (\frac{1}{2},\frac{\pi}{6}),(\frac{1}{2},\frac{5\pi}{6}),(\frac{1}{2},\frac{7\pi}{6}),(\frac{1}{2},\frac{11\pi}{6}).\)

However, an identical graph to \(\displaystyle r=\frac{1}{2}\) in polar coordinates is \(\displaystyle r=-\frac{1}{2}\), since these two equations describe the same circle with a radius \(\displaystyle \frac{1}{2}\) units long. Therefore, the equations \(\displaystyle r=cos(2\Theta)\) and \(\displaystyle r=-\frac{1}{2}\) must also be solved to yield the remaining points of intersection:

\(\displaystyle cos(2\Theta)=-\frac{1}{2}\),

\(\displaystyle \Rightarrow 2\Theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}.\)

\(\displaystyle \Rightarrow \Theta=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.\)

From this, it can be deduced that there are four other points of intersection between the given curves:

\(\displaystyle (-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{4\pi}{3}),(-\frac{1}{2},\frac{5\pi}{3}).\)

Hence, there are eight total points of intersection between the curves\(\displaystyle r=cos(2\Theta)\) and \(\displaystyle r=\frac{1}{2}\).

we are given \(\displaystyle r=-3cos(\theta)\)  and we know that

\(\displaystyle x=rcos(\theta )\)

\(\displaystyle y=rsin(\theta )\)

\(\displaystyle r^2=x^2+y^2\)

we have r and 3cos(theta). Multiplying each side of the equation by r would give us 

\(\displaystyle r^2=-3rcos(\theta )\)

substitute out the parts we know from the formulas above

\(\displaystyle x^2+y^2=-3x\)