How do I solve parametric and polar curve problems in Calculus 2.1?

[Jboeding]

Hey Everyone!

I have three questions that I do not know how to approach/solve. I've been checking online, the textbook, etc, and nothing. This is Calculus 2.1. Find the points with the given slope.
x=9cos(theta), y=9sin(theta), slope = 1/2.

Answer: (-9rt5/5, 18rt5/5), (9rt5/5, -18rt5/5)

Notes: My idea would be to set them equal, solve for theta, getting (pi/4) and (5pi/4). Next step: no idea.2. Write this is cartesian:
r = (4)/(3cos(theta) - 5 sin (theta))

Notes: I've never done a problem like this, and haven't found a similar one in the book or online, so I have no clue how to do this.

Answer: y = (4/5) - (3/5)x3. Find the area between the circle r=9 and the cartioid r=9(1-cos(theta))

Answer: ((405pi)/4) - 162

Notes: I set them equal to each other, got theta = to 0 and pi. But, the points of intersection are going to be (pi/2) and (3pi/2). I use the intersection points for my bounds in the formula: (1/2)Integral ((r2)^2-(r1)^2) from -pi/2 to pi/2.I've been stuck on these 3 problems for HOURS.
Any input, whether it is tips or the whole problem is greatly appreciated.

Thanks for the help everyone.
- Jacob

 

[soroban]

Hello, Jboeding!

2. Write this in cartesian: .[tex]r \:=\:\frac{4}{3\cos\theta - 5\sin\theta}[/tex]

[tex]\text{Clear the denominator: }\;3\underbrace{r\cos\theta}_{\text{This is }x} - 5\underbrace{r\sin\theta}_{\text{This is }y}\;=\;4[/tex]

[tex]\text{Therefore: }\;3x - 5y \:=\:4[/tex]

 

[Euge]

Hi Jacob,

I'll give you tips for all three problems.

Hey Everyone!

I have three questions that I do not know how to approach/solve. I've been checking online, the textbook, etc, and nothing. This is Calculus 2.1. Find the points with the given slope.
x=9cos(theta), y=9sin(theta), slope = 1/2.

Answer: (-9rt5/5, 18rt5/5), (9rt5/5, -18rt5/5)

Notes: My idea would be to set them equal, solve for theta, getting (pi/4) and (5pi/4). Next step: no idea.

The slope of the curve at $(x, y)$ is

$\displaystyle \frac{dy}{dx} = \dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{9\cos{\theta}}{-9\sin{\theta}} = -\cot{\theta}$

Since the slope is also $\frac{1}{2}$ at $(x, y)$, we have $\cot{\theta} = -\frac{1}{2}$. Drawing a reference triangle if necessary, we see that either $\cos{\theta} = \frac{1}{\sqrt{5}}$ and $\sin{\theta} = -\frac{2}{\sqrt{5}}$ or $\cos{\theta} = -\frac{1}{\sqrt{5}}$ and $\sin{\theta} = \frac{2}{\sqrt{5}}$. We get the two answers written by plugging these values in and rationalizing denominators.

2. Write this is cartesian:
r = (4)/(3cos(theta) - 5 sin (theta))

Notes: I've never done a problem like this, and haven't found a similar one in the book or online, so I have no clue how to do this.

Answer: y = (4/5) - (3/5)x

To go from polar to Cartesian, use $\frac{x}{r} = \cos{\theta}$ and $\frac{y}{r} = \sin{\theta}$.

3. Find the area between the circle r=9 and the cartioid r=9(1-cos(theta))

Answer: ((405pi)/4) - 162

Notes: I set them equal to each other, got theta = to 0 and pi. But, the points of intersection are going to be (pi/2) and (3pi/2). I use the intersection points for my bounds in the formula: (1/2)Integral ((r2)^2-(r1)^2) from -pi/2 to pi/2.

I don't think the given answer is right: it should be $162 - \frac{81\pi}{4}$. When you set them equal to each other, you end up with the equation $\cos{\theta} = 0$. So $\theta = \frac{\pi}{2}$ or $\theta = -\frac{\pi}{2}$. These give the limits of integration. By drawing the two curves, we see that the circle $r = 9$ is the outer curve and the cardioid $r = 9 (1 - \cos{\theta})$ is the inner curve. So the area of the region between these curves is

$\displaystyle \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 81(1-(1-\cos{\theta})^2)\, d\theta$,

which turns out to be $162 - \frac{81\pi}{4}$.

 

[Deveno]

I don't think the given answer is right: it should be $162 - \frac{81\pi}{4}$. When you set them equal to each other, you end up with the equation $\cos{\theta} = 0$. So $\theta = \frac{\pi}{2}$ or $\theta = -\frac{\pi}{2}$. These give the limits of integration. By drawing the two curves, we see that the circle $r = 9$ is the outer curve and the cardioid $r = 9 (1 - \cos{\theta})$ is the inner curve. So the area of the region between these curves is

$\displaystyle \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 81(1-(1-\cos{\theta})^2)\, d\theta$,

which turns out to be $162 - \frac{81\pi}{4}$.

I think the problem is poorly worded, perhaps by "between" they mean the area bounded by the "tighter" of the two curves, which in this case would be:

$\displaystyle \dfrac{81}{2}\left(\int_{-\pi/2}^{\pi/2} (1 - \cos\theta)^2\ d\theta + \int_{\pi/2}^{3\pi/2} d\theta\right)$ <--edited after the fact

in other words, the INTERSECTION of the two regions.

 

[Jboeding]

I think the problem is poorly worded, perhaps by "between" they mean the area bounded by the "tighter" of the two curves, which in this case would be:

$\displaystyle 81\left(\int_{-\pi/2}^{\pi/2} (1 - \cos\theta)^2\ d\theta + \int_{\pi/2}^{3\pi/2} d\theta\right)$

in other words, the INTERSECTION of the two regions.

Hey,
Why are both integrals multiplied by (81/2) and not just the first integral of where the 9^2 originated from?

But yeah, the answer is correct.
Big thanks :D

- Jacob

- - - Updated - - -

Hello, Jboeding!


[tex]\text{Clear the denominator: }\;3\underbrace{r\cos\theta}_{\text{This is }x} - 5\underbrace{r\sin\theta}_{\text{This is }y}\;=\;4[/tex]

[tex]\text{Therefore: }\;3x - 5y \:=\:4[/tex]

Hey,

In case you tried solving the problem, and couldn't get the answer supplied, I wrote the question down wrong. It was a (+) in the denominator instead of the (-).

Thanks for the help!
- Jacob

 

[Jboeding]

Hi Jacob,

I'll give you tips for all three problems.
The slope of the curve at $(x, y)$ is

$\displaystyle \frac{dy}{dx} = \dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{9\cos{\theta}}{9\sin{\theta}} = -\cot{\theta}$

Since the slope is also $\frac{1}{2}$ at $(x, y)$, we have $\cot{\theta} = -\frac{1}{2}$. Drawing a reference triangle if necessary, we see that either $\cos{\theta} = \frac{1}{\sqrt{5}}$ and $\sin{\theta} = -\frac{2}{\sqrt{5}}$ or $\cos{\theta} = -\frac{1}{\sqrt{5}}$ and $\sin{\theta} = \frac{2}{\sqrt{5}}$. We get the two answers written by plugging these values in and rationalizing denominators.
To go from polar to Cartesian, use $\frac{x}{r} = \cos{\theta}$ and $\frac{y}{r} = \sin{\theta}$.
I don't think the given answer is right: it should be $162 - \frac{81\pi}{4}$. When you set them equal to each other, you end up with the equation $\cos{\theta} = 0$. So $\theta = \frac{\pi}{2}$ or $\theta = -\frac{\pi}{2}$. These give the limits of integration. By drawing the two curves, we see that the circle $r = 9$ is the outer curve and the cardioid $r = 9 (1 - \cos{\theta})$ is the inner curve. So the area of the region between these curves is

$\displaystyle \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 81(1-(1-\cos{\theta})^2)\, d\theta$,

which turns out to be $162 - \frac{81\pi}{4}$.

Thank you for your thorough reply to all my questions!
I now have a better understanding, your answer helped a lot!
Question 3 was very difficult to understand.

For clarification (and to ease tension/questions, haha)... Here is the exact question:
"Find the area of the region common to the circle and the cardioid."
Question was solved in the post by another member if you are curious.

Thanks again for the help, I really appreciate it.
- Jacob

 

[Euge]

I think the problem is poorly worded, perhaps by "between" they mean the area bounded by the "tighter" of the two curves, which in this case would be:

$\displaystyle 81\left(\int_{-\pi/2}^{\pi/2} (1 - \cos\theta)^2\ d\theta + \int_{\pi/2}^{3\pi/2} d\theta\right)$

in other words, the INTERSECTION of the two regions.

Yes, I was thinking the question was asking for the area of the region that lies inside the circle $r = 9$ and outside the cardioid $r = 9(1-\cos\theta)$. But your answer turns out to be $\frac{405\pi}{2} - 324$, which is twice answer given by Jacob. I believe the area should be represented instead as

$81(\int_0^{\pi/2} (1-\cos\theta)^2\, d\theta + \int_{\pi/2}^\pi \, d\theta)$.

This does give Jacob's answer.

 

[Jboeding]

Yes, I was thinking the question was asking for the area of the region that lies inside the circle $r = 9$ and outside the cardioid $r = 9(1-\cos\theta)$. But your answer turns out to be $\frac{405\pi}{2} - 324$, which is twice answer given by Jacob. I believe the area should be represented instead as

$81(\int_0^{\pi/2} (1-\cos\theta)^2\, d\theta + \int_{\pi/2}^\pi \, d\theta)$.

This does give Jacob's answer.

Yeah, except that is it 81/2. I still do not quite understand why you multiply both integrals by that.

- Jacob

 

[Euge]

Yeah, except that is it 81/2. I still do not quite understand why you multiply both integrals by that.

- Jacob

I used the fact that the region is symmetric with respect to the x-axis. So the area you want is twice the area of the portion of the region above the x-axis. That's why you see the constant 81, rather than 81/2. The area of the region above the x-axis is

$\displaystyle \frac{1}{2} \int_0^{\pi/2} 81(1-\cos\theta)^2\, d\theta + \frac{1}{2} \int_{\pi/2}^\pi 81\, d\theta$.

The total area is twice this number, so the 1/2 goes away, giving the expression I had written.

 

[Deveno]

Yes, I was thinking the question was asking for the area of the region that lies inside the circle $r = 9$ and outside the cardioid $r = 9(1-\cos\theta)$. But your answer turns out to be $\frac{405\pi}{2} - 324$, which is twice answer given by Jacob. I believe the area should be represented instead as

$81(\int_0^{\pi/2} (1-\cos\theta)^2\, d\theta + \int_{\pi/2}^\pi \, d\theta)$.

This does give Jacob's answer.

I forgot the factor of 1/2 in front. The limits of integration are what they should be.

 

[Deveno]

Yeah, except that is it 81/2. I still do not quite understand why you multiply both integrals by that.

- Jacob

For a polar curve:

$r = f(\theta)$ the integral of the sector swept out from $\theta = a$ to $\theta = b$, is:

$\displaystyle \dfrac{1}{2}\int_a^b r^2\ d\theta=\dfrac{1}{2} \int_a^b (f(\theta))^2\ d\theta$

Basically, instead of summing a bunch of RECTANGLES of height $f(x)$ and width $dx$, we are summing a bunch of "pie slices" of radius $r = f(\theta)$, each of which has area:

$\pi r^2\cdot \dfrac{d\theta}{2\pi} = \dfrac{1}{2}r^2\ d\theta$ (the $\pi$'s cancel).

In this case, both functions are of the form:

$f(\theta) = 9\cdot\text{something}$

so when we square we get a constant factor of 81, that we can put "out front" with the 1/2, to get 81/2.

 

[Jboeding]

I forgot the factor of 1/2 in front. The limits of integration are what they should be.

I see, thank you so much. It's funny how the other way still gets the same answer though, haha.

- Jacob