Please help me check these answers

Mar 2010
15
0
Please help me check these answers (I)

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
Jun 2008
16,216
6,764
North Texas
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.
...
 
Mar 2010
15
0
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
Jun 2008
16,216
6,764
North Texas
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?

how about taking a derivative?

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.​

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

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

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

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

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

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

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

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

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

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

\(\displaystyle A = -\frac{12\pi}{5} \left[(4-x^\frac{2}{3})^\frac{5}{2}\right]_2^3 \approx 29.658\)
 
Mar 2010
15
0
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
Jun 2008
16,216
6,764
North Texas
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?
 
Mar 2010
15
0
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
Aug 2009
976
387
Israel
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...
 
Mar 2010
15
0
I'm sorry but I'm really lost and I have no idea what to do with that problem.