# I Need to verify these answers (surface area & parametrization)

• Feb 5th 2010, 09:14 PM
Shananay
I Need to verify these answers (surface area & parametrization)
I know this is a lot of stuff, but I want to be certain that I have all of this information correct. If there are any errors, I feel like they would come from the first problem or from $\frac{d^2y}{dx^2}$ on the last one. Many thanks to anyone willing to help me out with any of this.

1) Find the surface area of the solid obtained by revolving $\frac{1}{4}x = \sqrt{1-y}$ for $-1\leq y \leq 0$ about the y-axis

$x = 4\sqrt{1-y}$

$\frac{dx}{dy} = \frac{2}{\sqrt{1-y}}$

Surface Area $= 2\pi\int_{-1}^{0}4\sqrt{1-y}\left(\frac{2}{\sqrt{1-y}}\right)^2dy$

$= 8\pi\int_{-1}^{0}\sqrt{1-y}\left(\frac{4}{1-y}\right)dy$

$= 32\pi\int_{-1}^{0}\frac{\sqrt{1-y}}{1-y} dy$

$= 32\pi\int_{-1}^{0}\frac{1}{\sqrt{1-y}}dy$

$= -32\pi\left.(2\sqrt{1-y})\right|_{-1}^{0}$

$= -64\pi\left.(\sqrt{1-y})\right|_{-1}^{0}$

$= -64\pi-(-64\pi\sqrt{2})$

$= 64\pi(\sqrt{2}-1)$

2) Find an equation relating $x$ and $y$. Then sketch the curve C whose parametric equations are given and indicate the direction as $t$ increases

a) $x = 2t-1$, $y=8t^3-1$ for all t

$t=\frac{x+1}{2}$

$y=8\left(\frac{x+1}{2}\right)^3-1$

$y=(x+1)^3-1$

$y=x^3+3x^2+3x+1-1$

$y=x^3+3x^2+3x$

My graph looks like an elongated $y=x^3$ graph which goes through (0,0) and has a horizontal tangent at (-1,-1). The direction of t is going upwards.

b) $x=sin^2t,$ $y=cos t$ for $0 \leq t \leq \pi$

$x + y^2 = 1$

$y = \sqrt{1-x}$ not $\pm$ since $0 \leq x \leq 1$

My graph for this one is the right half of a circle of radius 1 centered on the origin. The direction of t is going clockwise.

3) Let C be the curve with parametrization $x=12t-t^3,$ $y=t^2-4$ for all t
Find equations of the tangent lines to C at the points corresponding to $t=1$ and $t=2$. At what point(s) will the graph have a horizontal tangent line? Also find $\frac{d^2y}{dx^2}$. Use all this information to sketch the curve C.

$\frac{dy}{dx}= \frac{2t}{12-3t^2}$

for $t=1$, $m=\frac{2}{9}$ and equation of the tangent line is $y+3=\frac{2}{9}(x-11)$

for $t=2$, m is undefined and the equation of the tangent line is $x=-16$

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dx}[2t(12-3t^2)^{-1}]}{12-3t^2}$

$= \frac{(2t)(-1)(12-3t^2)^{-2}(-6t)+2(12-3t^2)^{-1}}{12-3t^2}$

$= \frac{2(12-3t^2)^{-1}[1+6t^2(12-3t^2)^{-1}]}{12-3t^2}$

$= \frac{2[1+6t^2(12-3t^2)^{-1}]}{(12-3t^2)^2}$

$= \frac{2[1+\frac{6t^2}{12-3t^2}]}{(12-3t^2)^2}$

For the graph I have vertical tangents and x intercepts at $t=\pm2$ (-16, 0), (16, 0), a horizontal tangent at $t=0$ (0, -4). The graph crosses itself at the y-intercept (0, 8), is concave down while above the x-axis and concave up when below. It makes a big loop from left to right.

Any help is enormously appreciated (Wink)(Clapping)
• Feb 5th 2010, 11:48 PM
VonNemo19
Quote:

Originally Posted by Shananay

Any help is enormously appreciated (Wink)(Clapping)

You don't need any help. I didn't check the first problem because (I'll be honest) I didn't feel like it. But the rest of what you have looks good.

EDIT: Should 1. be $2\pi\int_{-1}^04\sqrt{1-y}\sqrt{1+[2(1-y)^{-1/2}]^2}dy=2\pi\int_{-1}^04\sqrt{1-y}\sqrt{1+4(1-y)^{-1}}dy$

$2\pi\int_{-1}^04\sqrt{(1-y)[1+4(1-y)^{-1}]}dy=2\pi\int_{-1}^04(\sqrt{1-y+4})dy$ ?
• Feb 6th 2010, 12:00 AM
Shananay
Quote:

Originally Posted by VonNemo19
You don't need any help. I didn't check the first problem because (I'll be honest) I didn't feel like it. But the rest of what you have looks good.

But I do need help:)

I need to verify my answers with someone else, and I can't do that myself. I'm one to always make careless errors and I need these answers to be perfect. Thank you for looking them over.
• Feb 6th 2010, 12:09 AM
VonNemo19
Quote:

Originally Posted by Shananay
But I do need help:)

I need to verify my answers with someone else, and I can't do that myself. I'm one to always make careless errors and I need these answers to be perfect. Thank you for looking them over.

As I said, your answers are fine. Spot on. Good work. Check out my last post. I've edited.
• Feb 6th 2010, 12:30 AM
Shananay
Quote:

Originally Posted by VonNemo19
Check out my last post. I've edited.

Thank you. I can't believe I messed up that formula. I'll have to re-evaluate that integral.
• Feb 6th 2010, 12:37 AM
VonNemo19
Quote:

Originally Posted by Shananay
Thank you. I can't believe I messed up that formula. I'll have to re-evaluate that integral.

...and just for your future benefit: When you were eliminating the paramater in 2. I would have stopped at $y=(x+1)^3-1$ because there is no reason whatsoever to go any further. Note that this is an excelent form to have this particular equation in because it is clear that this is nothing more than the graph of a cubic that has shifted one unit to the left and one unit down.
• Feb 6th 2010, 06:22 AM
Shananay
Quote:

Originally Posted by VonNemo19
When you were eliminating the paramater in 2. I would have stopped at $y=(x+1)^3-1$ because there is no reason whatsoever to go any further.

Good point, I didn't see that. It is much easier to graph that.