Help plz: My Final exam in Calculus III

[mooeypoo]

Hey guys,

 

After a VERY frustrating semester (work was tough, professor wasn't very helpful, and very big workload), I got a B+ in my math course. I don't like it at all, but it's better than failing, so I'm good.

 

I have the exam sheet still, and I would REALLY want to know where I screwed up. I have scribbled the answered that I put up in the actual exam on the edges of the questions, and I'm probably going to post lots of mistakes (it was stressing ) -- but I would like to know the answers.

 

I know it's not something we usually do here, solve questions, usually we ask to give the WAY -- well, feel free to give me a way too. I am going to post the questions and my scribbled solution. I can't remember what way I used, but I'll try.

 

If my answer's wrong, please show me the right way to do it... it's too late to change my exam obviously - it's only for my own perverted self-examination needs...

 

Thanks in advance

 

 

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Part I

 

1.

(a) Find an equation for the tangent plane to the surface S given by [math]z=8-x^2-3x^2[/math] at the point (2,1,1)

My answer: z-1=-4(x-2)-6(y-1)

 

(b) The plane x=2 intersects the surface S in a curve. Find a unit vector tangent to this curve at the point (2,1,1).

I think I overdid my divisions here.. anyways, my answer: [math]u=(\frac{2}{\sqrt{40}}, \frac{-6}{\sqrt{40}})[/math]

 

2. Consider the line: [math]r(t) = (4+2t)i + (2-t)j +(3-t)k[/math]

(a) Find the intersection point of that line with the plane z=4+x+3y or show that they do not intersect.

I don't have my answer written here and I don't remember off the top of my head what I wrote, but the way I remember it, I rewrote the line equation to x=4+2t, y=2-t, z=3-t and then plugged these values into the plane equation to see that it doesn't solve (hence, there are no intersecting lines). Is that right?

 

(b) Find an equation for the plane which contains that line and is perpendicular to the plane z=4+x+3y.

My answer:

x=4+4t

y=2+t

z=3+7t

And so the n (normal vector) = n x v = (4,1,7)

(v is the line vector, n is the normal vector to the plane)

 

3. Evaluate:

[math]\int\int_{T}(x)dA[/math] where T is the triangle with vertices at the origin, (1,1) and (2,0).

 

My answer:

[math]\int^{1}_{0}\int^{2-y}_{y}(x)dxdy = 1[/math]

 

4. Let [math]f(x,y)=xy^2-2x^2[/math]

(a) Find the directional derivative of f at the point (2,-1) in the direction of v=3i-4j

My Answer: [math]u\nabla f=16/5[/math]

 

(b) Give a unit vector in the direction of maximum decrease of f(x,y) at (2,-1)

I was quite proud of myself of noticing it said maximum DEcrease (which is actually the "minimum" increase.. ha. But I'm not sure I got the answer right

My answer: [math](\frac{7}{\sqrt{65}},\frac{4}{\sqrt{65}})[/math]

 

5. For each of the following series, state whether the series is absolutely convergent, conditionally convergent, or divergent. Credit will not be given unless the reasons for your conclusions are given.

(Note: I can't stand series.. I get confused. It probably shows)

(a) [math]\sum^{\infty}_{n=1}\frac{(-1)^n}{2n+3}[/math]

Conditionally convergent.

 

(b) [math]\sum^{\infty}_{n=2}\frac{(-1)^n}{\sqrt{n}ln(n)}[/math]

Divergent.

 

© [math]\sum^{\infty}_{n=2}\frac{(2n-5)}{n^3-4n+3}[/math]

Absolutely convergent.

 

6. Find the interval of convergence for the power series [math]\sum^{\infty}_{n=1}\frac{(2x-5)^n}{n+1}[/math] (remember to check the endpoints)

My Answer: [2,3)

 

7. Evaluate the triple integral [math]\int\int\int_{G}ZdV[/math] where G is the region between the spheres x^2+y^2+z^2=1 and x^2+y^2+z^2=4 in the first octant.

I don't have a written answer to this one, don't remember what I wrote eventually, but I overdid this one completely.. I got really confused. I have semi-answer (beginning of my methods):

[math]\int^{\pi/2}_{0}\int^{2}_{1}\int^{\sqrt{4-r^2}}_{\sqrt{1-r^2}}zr dzdrd\theta[/math]

 

Part II (Answered 3 of 5 questions):

8. Find and classify the critical points of [math]f(x,y)=xy^2+8x^2+y^2[/math]

My answer:

(-1,4) and (-1,-4) --> Saddle points.

(0,0) --> Minimum point.

 

10. Sketch the region of integration, change the order of integration and evaluate:

[math]\int^{9}_{0}\int^{\sqrt{x}}_{x/3}2x dydx[/math]

My Answer (minus the sketch):

[math]\int^{3}_{0}\int^{y^2}_{3y}2x dxdy[/math]

 

12. For each of the following, find the limit or show that the limit does not exist:

(a) [math]\lim_{x,y \to 0,0}\frac{x^2+xy+y^2}{x^2+y^2}[/math]

My answer:

path 1 --> x=0

path 2 --> y=0

Both limits are different ==> Limit does not exist.

 

(b) [math]\lim_{x,y \to 0,0}\frac{x^4-y^4}{x^2+y^2}[/math]

My answer: The limit DOES exist:

[math]\lim_{x,y \to 0,0}\frac{(x^2-y^2)(x^2+y^2)}{x^2+y^2}[/math]

[math]= \lim_{x,y \to 0,0}(x^2+y^2)=0[/math]

 

 

--

 

That's it. I will really appreciate any help in this. I'm curious to know what I got right and what I got wrong, and from the things I got wrong - how to do better...?

 

Unfortunately, we don't have answersheets for finals :\

 

Thanks in advance!

 

~moo

Edited June 17, 2008 by mooeypoo

[Dave]

If my answer's wrong, please show me the right way to do it... it's too late to change my exam obviously - it's only for my own perverted self-examination needs...

 

Thanks in advance

 

Unfortunately I don't have time to do the whole thing, but I've done a few of the questions and that should be something, at least. Hopefully somebody else can come and fill in the blanks.

 

3. Evaluate:

[math]\int\int_{T}(x)dA[/math] where T is the triangle with vertices at the origin, (1,1) and (2,0).

 

My answer:

[math]\int^{1}_{0}\int^{2-y}_{y}(x)dxdy = 1[/math]

 

I don't think this is correct, as the area you're actually integrating over isn't correct. However, I've been known to be wrong myself with these things. To be sure, split the triangle up into two parts, and then the answer should be:

 

[math]\int_{0}^{\frac{1}{2}} \int_{0}^{2x} x \, dy dx + \int_{\frac{1}{2}}^1 \int_{0}^{2-2x} x \, dy dx[/math]

 

4. Let [math]f(x,y)=xy^2-2x^2[/math]

(a) Find the directional derivative of f at the point (2,-1) in the direction of v=3i-4j

My Answer: [math]u\nabla f=16/5[/math]

 

[math]\nabla f(\vec{x}) = (y^2-4x, 2xy)[/math], and so

 

[math]\nabla_{(3,-4)} f(2, -1) = (3, -4) \cdot \nabla f(2, -1) = (3, -4) \cdot (-7, -4) = -21 + 16 = -5[/math]

 

5. For each of the following series, state whether the series is absolutely convergent, conditionally convergent, or divergent. Credit will not be given unless the reasons for your conclusions are given.

(Note: I can't stand series.. I get confused. It probably shows)

(a) [math]\sum^{\infty}_{n=1}\frac{(-1)^n}{2n+3}[/math]

Conditionally convergent.

 

(b) [math]\sum^{\infty}_{n=2}\frac{(-1)^n}{\sqrt{n}ln(n)}[/math]

Divergent.

 

© [math]\sum^{\infty}_{n=2}\frac{(2n-5)}{n^3-4n+3}[/math]

Absolutely convergent.

 

(a) is correct since [imath]\sum a_n[/imath] will converge by the alternating series test, but [imath]\sum |a_n|[/imath] is similar to the harmonic series and so will diverge.

 

(b) Not quite. It's the same behaviour as the series above, only in another guise. [math]\sqrt{n}[/math] and [math]\log(n)[/math] are both strictly increasing functions, so [math]a_n = 1/\sqrt{n}\log(n)[/math] will be a strictly decreasing sequence. Hence, by the alternating series test, it will converge. However, the absolute series is not convergent, since

 

[math]\log n \leq \sqrt{n} \Rightarrow \frac{1}{\log n} \geq \frac{1}{\sqrt{n}} \Rightarrow \frac{1}{\sqrt{n} \log n} \geq \frac{1}{n}[/math]

 

So, by the comparison test, [imath]\sum |a_n| \geq \sum \frac{1}{n}[/imath] and the series diverges.

 

© looks correct but I haven't checked it.

 

6. Find the interval of convergence for the power series [math]\sum^{\infty}_{n=1}\frac{(2x-5)^n}{n+1}[/math] (remember to check the endpoints)

My Answer: [2,3)

 

I get the same answer.

 

7. Evaluate the triple integral [math]\int\int\int_{G}ZdV[/math] where G is the region between the spheres x^2+y^2+z^2=1 and x^2+y^2+z^2=4 in the first octant.

I don't have a written answer to this one, don't remember what I wrote eventually, but I overdid this one completely.. I got really confused. I have semi-answer (beginning of my methods):

[math]\int^{\pi/2}_{0}\int^{2}_{1}\int^{\sqrt{4-r^2}}_{\sqrt{1-r^2}}zr dzdrd\theta[/math]

 

Yeah, it looks like you've tried to change into cylindrical polar co-ordinates when in fact spherical will be easier to work with. All you have to do is calculate the integrals over the two spheres and then subtract to get the region in-between. So, for the smaller sphere, I apply the transformation:

 

[math]x = r \sin\theta \cos\varphi, y = r\sin\theta \sin\varphi, z = r\cos\theta[/math]

 

and then evaluate:

 

[math]\int_{0}^{2\pi} \int_{0}^{\pi} \int_0^1 r\cos\theta \cdot r^2 \sin\theta dr d\varphi d\theta[/math]

 

Part II (Answered 3 of 5 questions):

8. Find and classify the critical points of [math]f(x,y)=xy^2+8x^2+y^2[/math]

My answer:

(-1,4) and (-1,-4) --> Saddle points.

(0,0) --> Minimum point.

 

Didn't check the stability but fixed points are correct.

 

10. Sketch the region of integration, change the order of integration and evaluate:

[math]\int^{9}_{0}\int^{\sqrt{x}}_{x/3}2x dydx[/math]

My Answer (minus the sketch):

[math]\int^{3}_{0}\int^{y^2}_{3y}2x dxdy[/math]

 

I think the answer is actually:

 

[math]\int^{3}_{0}\int^{3y}_{y^2}2x dxdy[/math]

 

Have a look at the plot and orient your head so that you can see what's going on.

 

12. For each of the following, find the limit or show that the limit does not exist:

(a) [math]\lim_{x,y \to 0,0}\frac{x^2+xy+y^2}{x^2+y^2}[/math]

My answer:

path 1 --> x=0

path 2 --> y=0

Both limits are different ==> Limit does not exist.

 

If you put x = 0, then your limit becomes

 

[math]\lim_{x,y \to 0,0}\frac{y^2}{y^2} = 1[/math]

 

Similarly if I put y = 0, then we get

 

[math]\lim_{x,y \to 0,0}\frac{x^2}{x^2} = 1[/math]

 

So these agree. However, if I put x = y, we get:

 

[math]\lim_{x,y \to 0,0}\frac{3x^2}{2x^2} = \frac{3}{2}[/math]

 

So you're correct, but your explanation isn't quite there.

 

(b) [math]\lim_{x,y \to 0,0}\frac{x^4-y^4}{x^2+y^2}[/math]

My answer: The limit DOES exist:

[math]\lim_{x,y \to 0,0}\frac{(x^2-y^2)(x^2+y^2)}{x^2+y^2}[/math]

[math]= \lim_{x,y \to 0,0}(x^2+y^2)=0[/math]

 

Looks good.