#### myngo9191

Hi, I'm working on some problems for reviewing, but I'm not sure if the answers I'm getting are the right one. Please help me check if they're right. If they're not please let me know and I will post my work to see what I have done wrong.

1. Find

2. Find the equation of the line tangent to at

3. Find the point P on the graph of closest to the point

4. Calculate the surface area of revolution of about the x-axis over

=

Last edited:

#### skeeter

MHF Helper
Hi, I'm working on some problems for reviewing, but I'm not sure if the answers I'm getting are the right one. Please help me check if they're right. If they're not please let me know and I will post my work to see what I have done wrong.

1. Find

ok

2. Find the equation of the line tangent to at

what is "b" ?

3. Find the point P on the graph of closest to the point

so far so good ... and the point is ... ?

4. Calculate the surface area of revolution of about the x-axis over

=

no.
...

#### myngo9191

2. Find the equation of the line tangent to at

$$\displaystyle y=\frac{5}{13}x+$$$$\displaystyle \frac{21}{13}$$
3. Find the point P on the graph of closest to the point

How do I find the point from here? Do I plug 10 in for x?

4. Calculate the surface area of revolution of about the x-axis over
=

Here is my work, please let me know where I went wrong.​

$$\displaystyle y=(4-x^\frac{2}{3})^\frac{3}{2}$$

$$\displaystyle y'=-x^\frac{-1}{3}(4-x^\frac{2}{3})^\frac{1}{2}$$

$$\displaystyle (y')^2+1\rightarrow 1+(4-x^\frac{2}{3})^\frac{1}{2})^2=1+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{x^\frac{2}{3}}{x^\frac{2}{3}}+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{4}{x^\frac{2}{3}}$$

$$\displaystyle u=4-x^\frac{2}{3}$$
$$\displaystyle du=-\frac{2}{3}x^{-\frac{1}{3}}dx$$
$$\displaystyle -3du=2x^{-\frac{1}{3}}$$

$$\displaystyle 2\pi\int_{2}^{3}u^{\frac{3}{2}}(-3)du=-6\pi\frac{2}{5}u^{\frac{5}{2}}\Bigg|_{2}^{3}=-\frac{12\pi}{5}(4-x^{\frac{2}{3}})^{\frac{5}{2}}\Bigg|_{2}^{3}$$

#### skeeter

MHF Helper
2. Find the equation of the line tangent to at

$$\displaystyle y=\frac{5}{13}x+$$$$\displaystyle \frac{21}{13}$$

ok
3. Find the point P on the graph of closest to the point

How do I find the point from here? Do I plug 10 in for x?

4. Calculate the surface area of revolution of about the x-axis over
=

Here is my work, please let me know where I went wrong.​

$$\displaystyle y=(4-x^\frac{2}{3})^\frac{3}{2}$$

$$\displaystyle y'=-x^\frac{-1}{3}(4-x^\frac{2}{3})^\frac{1}{2}$$

$$\displaystyle (y')^2+1\rightarrow 1+(4-x^\frac{2}{3})^\frac{1}{2})^2=1+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{x^\frac{2}{3}}{x^\frac{2}{3}}+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{4}{x^\frac{2}{3}}$$

$$\displaystyle u=4-x^\frac{2}{3}$$
$$\displaystyle du=-\frac{2}{3}x^{-\frac{1}{3}}dx$$
$$\displaystyle -3du=2x^{-\frac{1}{3}}$$

$$\displaystyle 2\pi\int_{2}^{3}u^{\frac{3}{2}}(-3)du=-6\pi\frac{2}{5}u^{\frac{5}{2}}\Bigg|_{2}^{3}=-\frac{12\pi}{5}(4-x^{\frac{2}{3}})^{\frac{5}{2}}\Bigg|_{2}^{3}$$
$$\displaystyle A = 2\pi \int_2^3 y\sqrt{1+(y')^2} \, dx$$

$$\displaystyle y = (4-x^\frac{2}{3})^\frac{3}{2}$$

$$\displaystyle y' = -\frac{(4-x^\frac{2}{3})^\frac{1}{2}}{x^\frac{1}{3}}$$

$$\displaystyle (y')^2 = \frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}$$

$$\displaystyle 1 + (y')^2 = \frac{4}{x^\frac{2}{3}}$$

$$\displaystyle A = 2\pi \int_2^3 (4-x^\frac{2}{3})^\frac{3}{2} \cdot \frac{2}{x^\frac{1}{3}} \, dx$$

$$\displaystyle A = -\frac{12\pi}{5} \left[(4-x^\frac{2}{3})^\frac{5}{2}\right]_2^3 \approx 29.658$$

#### myngo9191

3. Find the point P on the graph of closest to the point

I got this after taking the derivative

$$\displaystyle 2(x-10)+1$$

Now what do I do?

#### skeeter

MHF Helper
3. Find the point P on the graph of closest to the point

I got this after taking the derivative

$$\displaystyle 2(x-10)+1$$

Now what do I do?
have you not completed any optimization problems?

where do extrema occur?

#### myngo9191

I didn't even know it's an optimization problem and no I haven't done any. I went look in the book and it says I should find the critical point by solving for x. I'm not sure if that's the way to go but here it is...

$$\displaystyle 2(x-10)+1\rightarrow x=9\frac{1}{2}$$

#### Defunkt

MHF Hall of Honor
The point closest to the graph is the point where the minimum of the distance function is achieved.

Minimum is achieved at an extremal point, so...

#### myngo9191

I'm sorry but I'm really lost and I have no idea what to do with that problem.