Math problems to tell where you are with your knowledge of math

[2slowtogofast]

Sort of a strange topic But I was wondering if some of you could suggest some problems for me to try to see how much math I understand. My background is I am a Mech Eng student in my last year of my bachelors degree. So I have taken calc 1-3 Diff Eq, Linear and a Applied math course for engineers. I guess I just would like to get a good idea of what I know. I do problems online and out of textbooks but I don't really know how hard or easy they are compared to what other people do. I am not asking for impossible questions just some questions that some one with my background should be able to do that are interesting or challenging. Thanks

 

[micromass]

1) Find $\int \log(x)dx$.

2) Find $\int_0^\infty \int_0^\infty e^{-x^2-y^2}dxdy$ and deduce $\int_0^\infty e^{-x^2}dx$.

3) Does the series $\sum{\frac{n^2}{3^n}}$ converge?? Find its limit.

4) Find the Fourier series of the function f(x)=x if $x\in [0,\pi]$ and $f(x)=0$ if $x\in [-\pi,0]$.

5) Is the series $\sum{ \frac{1}{3^n}\sin(nx)}$ the Fourier series of some function?? Of which one??

6) Take the function $f(x)=\frac{1}{x^2}$ for x>1 and rotate it around the x-axis. What is it's volume and surface area??

7) Let $A=\left(\begin{array}{ccc} -1 & 0 & -1\\ -1 & 0 & 0\\ 2 & 1 & 2\\\end{array}\right)$. Find $A^{1000}$.

8) Solve the following system

$$\left\{\begin{array}{cc} u^\prime= 3u+4v\\ v^\prime= u+v\end{array}\right.$$

9) At time t=0 a tank contains Q₀kg of salt of salt dissolved in 100l of water. Assume that water containing 1/4 kg of salt/l is entering the tank at a rate of r l/min and that the well-stirred mixture is draining from the tank at the same rate. Find the limiting amount of salt that is present after a long time. After which time is the salt level within 2% of the limiting amount? What flow rate would be required to obtain a salt level within 2% if it is required that the time does not exceed 45 min.

Something more theoretical:

10) Classify all the 6x6 matrices A such that the column space of A equal the nullspace of A.

 

[2slowtogofast]

Thank you this was what I am looking for. I am going to start working through these tommorow and I will post my solutions and hopefully you guys can let me know if they are correct. Thanks

 

[2slowtogofast]

1.) I always forget if you write log without a base is that base e or 10? I went with e here is my solution

∫logx dx = uv - ∫v du

u = logx du = 1/x
dv = dx v = x

plug in


∫logx dx = xlogx - ∫x (1/x) dx

∫logx dx = xlogx - ∫ dx

∫logx dx = xlogx - x

 

[micromass]

Correct, but you forgot the integration constant...

 

[2slowtogofast]

2.)

$\int^{∞}_{0}\int^{∞}_{0}$e^-x2-y2dxdy

Change to Polar Co-ords

-x²-y²=r²cos$\theta$²-r²sin$\theta$²

Factor the r² and sin²+cos² goes to 1

-x²-y²=-r²

Rewrite the integral


$\int^{∞}_{0}\int^{\pi}_{0}$e^-r2rd$\theta$dr

Since there is no theta everything is treated as a constant Evaluating the inner intergral


$\pi$$\int^{∞}_{0}$e^-r2rdr

U substitution: u = r² After finding du and solving for dx then plugging in we get

$\frac{\pi}{2}$$\int^{∞}_{0}$e^-udr

=$\frac{\pi}{2}$(-e^-∞+e⁰)

=$\frac{\pi}{2}$

I skipped a few steps when typing this because i am not very familar with latex. This problem took a while but i think I got it spent a lot of time going through my calc book. For the second part is it just half of the answer so pi/4 because it is the same integral with the same limits repeated twice?

Anyway I have a question on the next one let me know if i am on the right track. I used the integral test to show that it converges but i can not find what it goes to. I tried taking the lim as n goes to inf and then using l'hospitals rule but no luck.

 

[micromass]

2.)

$\int^{∞}_{0}\int^{∞}_{0}$e^-x2-y2dxdy

Change to Polar Co-ords

-x²-y²=r²cos$\theta$²-r²sin$\theta$²

Factor the r² and sin²+cos² goes to 1

-x²-y²=-r²

Rewrite the integral


$\int^{∞}_{0}\int^{\pi}_{0}$e^-r2rd$\theta$dr

The inner integral should go from 0 to $\pi/2$.

Since there is no theta everything is treated as a constant Evaluating the inner intergral


$\pi$$\int^{∞}_{0}$e^-r2rdr

U substitution: u = r² After finding du and solving for dx then plugging in we get

$\frac{\pi}{2}$$\int^{∞}_{0}$e^-udr

=$\frac{\pi}{2}$(-e^-∞+e⁰)

=$\frac{\pi}{2}$

I skipped a few steps when typing this because i am not very familar with latex. This problem took a while but i think I got it spent a lot of time going through my calc book. For the second part is it just half of the answer so pi/4 because it is the same integral with the same limits repeated twice?

It's not half the answer but rather the square root of the answer. Do you see why??

Anyway I have a question on the next one let me know if i am on the right track. I used the integral test to show that it converges but i can not find what it goes to. I tried taking the lim as n goes to inf and then using l'hospitals rule but no luck.

The trick is to start from

$$\frac{1}{1-x}=\sum {x^n}$$

and to differentiate both sides.

 

[2slowtogofast]

I see why its sqd now b/c you can write it as (e^-x2)(e^-y2) so since the limits are the same the only thing different is the x and y should have seen that.

I don't see why the limit is pi/2 though if the original int was -inf to inf then it would be 0 to 2pi right? so if that is true would't this case be 0 to pi because the integral goes from 0 to inf. I am on my phone so i couldt type this out in latex sorry

 

[micromass]

I see why its sqd now b/c you can write it as (e^-x2)(e^-y2) so since the limits are the same the only thing different is the x and y should have seen that.

I don't see why the limit is pi/2 though if the original int was -inf to inf then it would be 0 to 2pi right? so if that is true would't this case be 0 to pi because the integral goes from 0 to inf. I am on my phone so i couldt type this out in latex sorry

The area that you integrate over is the first quadrant. If you want to cover that area with polar coordinates, then your angle can only go from 0 to pi/2. An angle higher than pi/2 would go out of the first quadrant.

 

[2slowtogofast]

I am taking a break from 3 here is 8.

Solve the system:

u' = 3u + 4v
v' = u + v

Solve v' for u

u = v' - v

Differentiate both sides

u' = v'' - v'

Sub u and u' into the first eqn to get an eqn only in v

v'' - v' = 3(v' - v) + 4v

This will simplify to

v'' - 4v' - 1 = 0
Write the C.E.
s² - 4s - 1 = 0

The roots are 2+/- √(5)

v(t) = C₁e^{-(2- √(5)t)} + C₂e^{-(2+ √(5)t)}

Take the derivative of v(t)

v'(t) = -(2 - √(5))C₁e^{-(2- √(5)t)} - (2+ √(5))C₂e^{-(2+ √(5)t)}

Use u = v' - v to solve fot u(t)

After substitution of v' and v we get

u(t) = (-3 + √(5) C₁e^{-(2- √(5)t)} + (-3 - √(5) )C₂e^{-(2+ √(5)t)}

Since no I.C.'s can not solve for constants