# Vectors and Integrals

1. Sep 4, 2005

### lynn32

Hi, I have a few summer assignment questions, if someone could help me taht would be great.

1. What are the units of a unit vector?
-I think that a unit vector has no units or is the question talking about a different type of units. I'm just confused.

2.Evaluate the following definite integrals
5
ò x–1 dx= ln x +C
0

ln5 – ln0

ln(0) is undefined so I am unsure what to do from here on this problem.

3. Express in terms of ln 2 and ln 5:

(b) ln (1/4)

(d) ln (1/40)

I don’t know what to do with the fractions in this problem. I did this one with no problem.
ln(50)= ln2 + 2ln5

4. Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be one hundred times larger than the magnitude of A – B, what must be the angle between them?
I really have no idea how to go about doing this problem. Maybe if someone could point me in the right direction or give me a hint about where to start I would be able to figure something out.

Thanks so much.
-Lynn

2. Sep 4, 2005

### TD

1) A unit vector (or vector in general) doesn't have a 'unit' as far as I know, but perhaps something else is meant.

2) I assume that -1 is the power?
Since ln(x) isn't defined in 0, you take the limit there and just calculate it like you always do.

$$\int\limits_0^5 {\frac{1}{x}dx} = \left[ {\ln x} \right]_0^5 = \ln 5 - \mathop {\lim }\limits_{t \to 0} \left( {\ln t} \right) = \ln 5 - \left( { - \infty } \right) = \infty$$

So the integral is divergent.

3)

(b) $$\ln \left( {\frac{1}{4}} \right) = \ln \left( {\frac{1}{{2^2 }}} \right)$$

(d) 40 can be written as 8*5, and 8 is 2^3...

3. Sep 5, 2005

### andrevdh

lynn32, you are looking for the angle phi.

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4. Sep 5, 2005

### HallsofIvy

Staff Emeritus
As stated, the question makes no sense. Is that all there was to it? Perhaps there was more. Is it possible that the question was asking what the word "unit" in "unit vector" meant?

If this really was $$\int_0^5 x^{-1}dx$$ then the answer is "undefined". If it is really $$\int_0^5 (x-1)dx$$, which is what you wrote, then it's much simpler!

ln(a/b)= ln(a)- ln(b).

The length of A-B is (A-B).(A-B)= |A|2- 2A.B+ |B|2= 2|A|2- |A|2cos(θ) (since |A|= |B| ) where &theta; is the angle between A and B.
The length of A+ B is (A+B).(A+B)= |A|2+2A.B+|B|2= 2|A|2+ A2cos(&theta;).

If "the magnitude of A + B to be one hundred times larger than the magnitude of A – B" then
2|A|2+ 2|A|2cos(θ)= 10(2|A|2- 2|A|2cos(θ)) .

Solve that for θ. Notice how nicely the "2|A|2" terms cancel!

Last edited: Sep 5, 2005
5. Sep 5, 2005

### TD

Really? I thought these kind of improper integrals could be solved as I described since some may converge and others would diverge (such as the on here). So the answer isn't inf. then?

6. Sep 5, 2005

### lynn32

It was $$\int_0^5 x^{-1}dx$$ this integral and not the simpler one.

For the vector question, how did you get from
(A-B).(A-B)= |A|2- 2A.B+ |B|2 to 2|A|2- |A|2cos(θ)
I don't understand where the cosine comes from.

On a side note, HallsofIvy, I noticed that you teach at Gallaudet University. I am in my third year of ASL in my high school and love it. My teacher is hearing but has a Deaf sister and attended Gallaudet to become an interpreter.

Thanks for taking the time to help me.
-Lynn

Last edited: Sep 5, 2005
7. Sep 5, 2005

### lynn32

Can I ask a couple more questions? These are giving me some trouble.

(d) What is the magnitude of A x (B x A) if there is an angle $$\theta$$ between A and B?

Is this enough of answer or do I need to go further.
$$B \times A=BAsin \theta$$
$$A \times (B \times A)= A(ABsin \thetasin \theta)$$
$$A \times (B \times A)=A^2Bsin^2 \theta$$

7. Let two vectors be represented in their coordinates as
A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk
(a) Show that A · B = AxBx + AyBy + AzBz (show all steps in this proof)

This answer also seems to simple to be correct:
A = (Ax, Ay, Az)
B = (Bx, By, Bz)
A · B = AxBx + AyBy + AzBz

Thanks a lot.
Lynn

8. Sep 5, 2005

### LeonhardEuler

Be careful about which quantities are vectors and which are scalars. BxA is a vector whose magnitude is $AB\sin{\theta}$. That does not mean that $\vec{B}\times\vec{A}=AB\sin{\theta}$. Actually, $\vec{B}\times\vec{A}=AB\vec{u}\sin{\theta}$ where u is a vector of magnitude 1 perpendicular to both A and B. So the question is asking to take the cross product of this with A. Since u is perpendicular to A, this gives $A^2B\sin{\theta}\sin{90}=A^2B\sin{\theta}$.
Well, that depends on what you're book takes as the definition of the dot product. If it takes as a definition that (Ax,Ay,Az)·(Bx,By,Bz)=AxBx + AyBy + AzBz, then that's fine. But I suspect they take the definition to be $\vec{A}\cdot\vec{B}=AB\cos{\theta}$ in which case you will need to show that this is equivalent. Draw out the vectors and use the formulas for the cosine of a sum or difference of angles to get the proof.

Last edited: Sep 5, 2005
9. Sep 5, 2005

### Neohaven

integrals... is this in the wrong section, or does secondary math include differentiation and integration?

Also... when we are speaking about K-12, what is twelvth grade? College? cuz here in canada we dont have any 12th grade... maybe it's because some countries/states have 7th grades?

Last edited: Sep 5, 2005
10. Sep 6, 2005

### HallsofIvy

Staff Emeritus
From last to first:
Neohaven: in the United States, children start "first grade" at 6 years old. Typically "high school" is grades 9 to 12, normally 14 to 18 years old. Yes, some secondary school (high school) students take calculus, especially those taking "advanced placement" classes that count toward college. Frankly, secondary math is in disarray- SOME students take calculus in secondary school- some pretty good, some "cookbook" (memorize: (x2)'= 2x !!!), which in my opinion hurt more than they help learning REAL caclulus. Large numbers of students have to take remedial algebra courses in college.

lynn32:"(d) What is the magnitude of A x (B x A) if there is an angle between A and B?"
You need to remember that (BxA) is directed perpendicular to both B and A. The magnitude of BxA is |A||B|sin(θ). Since sin(π/2)= 1, The magnitude of Ax(BxA) is (|A||B|sin(&theta;))(|A|) (1)= |A|2|B|sin(&theta;).

"For the vector question, how did you get from
(A-B).(A-B)= |A|2- 2A.B+ |B|2 to 2|A|2- |A|2cos(θ)
I don't understand where the cosine comes from. "

Okay: (A-B).(A-B)= A.A- A.B- B.A+ B.B= |A|2- 2A.B+ |B|2 you understand. One way of defining A.B is |A||B|cos(θ) where θ is the angle between the vectors, which is what you were asked to find (surely if you know |AxB|= |A||B|sin(θ), you know that!). Since we are told that |A|= |B|, A.B= |A||B|cos(θ)= |A|2cos(θ) and |A|2+|B|2= 2|A|2.
(A-B).(A-B)= 2|A|2- 2|A|2cos(θ)
and, of course
(A+B).(A+B)= 2|A|2+ 2|A|2cos(θ)

TD: "Really? I thought these kind of improper integrals could be solved as I described since some may converge and others would diverge (such as the on here). So the answer isn't inf. then?"
Yes, some may converge and other diverge. This diverges: the integral does not exist. "infinity" is NOT a number. Saying that the integral is "infinity" is just a way of saying that the integral does not exist.

11. Sep 6, 2005

### cscott

I live in Canada and I begin the twelvth grade tomorrow . College/University is after the twelvth grade.