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

=

2. Originally Posted by 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

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.
...

3. 2. Find the equation of the line tangent to at
$\displaystyle y=\frac{5}{13}x+$$\displaystyle \frac{21}{13} 3. Find the point P on the graph of closest to the point How do I find the point from here? Do I plug 10 in for x? 4. Calculate the surface area of revolution of about the x-axis over = Here is my work, please let me know where I went wrong. \displaystyle y=(4-x^\frac{2}{3})^\frac{3}{2} \displaystyle y'=-x^\frac{-1}{3}(4-x^\frac{2}{3})^\frac{1}{2} \displaystyle (y')^2+1\rightarrow 1+(4-x^\frac{2}{3})^\frac{1}{2})^2=1+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{x^\frac{2}{3}} {x^\frac{2}{3}}+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{4}{x^\frac{2}{ 3}} \displaystyle u=4-x^\frac{2}{3} \displaystyle du=-\frac{2}{3}x^{-\frac{1}{3}}dx \displaystyle -3du=2x^{-\frac{1}{3}} \displaystyle 2\pi\int_{2}^{3}u^{\frac{3}{2}}(-3)du=-6\pi\frac{2}{5}u^{\frac{5}{2}}\Bigg|_{2}^{3}=-\frac{12\pi}{5}(4-x^{\frac{2}{3}})^{\frac{5}{2}}\Bigg|_{2}^{3} 4. Originally Posted by myngo9191 2. Find the equation of the line tangent to at \displaystyle y=\frac{5}{13}x+$$\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.

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

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

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

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

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

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

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

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

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

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

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

5. 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?

6. Originally Posted by 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?
have you not completed any optimization problems?

where do extrema occur?

7. 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...

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

8. 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...

9. go to the link ...

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxmindirectory/MaxMin.html

read the guidelines then have a look at problem #7.

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

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

$\displaystyle d^2(x)'=2(x-10)+1$

$\displaystyle \frac{-1}{2(x-10)+1} = (y-0)(x-10)$

$\displaystyle \Big\Downarrow$

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

$\displaystyle \Big\Downarrow$

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

$\displaystyle \Big\Downarrow$

$\displaystyle \frac{-1}{(2(x-10)+1)\sqrt{x}}-x+10=0$

$\displaystyle \Big\Downarrow$

$\displaystyle -1-x+10=0$

$\displaystyle \Big\Downarrow$

$\displaystyle x=9$

$\displaystyle \Big\Downarrow$

$\displaystyle y=\sqrt{x}=\sqrt{9}=3$

so the point on the graph of $\displaystyle y=\sqrt{x}$ closest to the point (10,0) is (9,3)
Please let me know if this is right and if not where I have messed up at. Thank you

12. $\displaystyle D^2 = (x-10)^2 + x$

$\displaystyle \frac{d(D^2)}{dx} = 2(x-10) + 1$

set the derivative equal to 0 because extrema occur at critical values ...

$\displaystyle 2(x-10) + 1 = 0$

solve for x ...

$\displaystyle x = 9.5$

$\displaystyle D^2$ has a minimum value at $\displaystyle x = 9.5$ ... you can use the 1st or 2nd derivative test(s) to confirm, or you can just note that the graph of $\displaystyle D^2$ is a standard parabola with vertex at the bottom.

so, the closest point on the curve $\displaystyle y = \sqrt{x}$ to the point $\displaystyle (10,0)$ is the point $\displaystyle \left(9.5,\sqrt{9.5} \right)$