Solving the Square Root of a Difference: Integrating √(9-x^2)

[Dickfore]

[STRIKE]What makes you think next time you will have any comments?[/STRIKE]

EDIT:

I just noticed you were not the person looking for the solution, but actually checked through their solution. My bad.

 

[mrigmaiden]

Jeges don't mind him just examine what he's said, as he clearly is lacking social skills! I have another way you can visualize how to find cos(sin^-1(x/3)) and it involves constructing your triangles similar to what you've already done. You know that for right triangles sin(x)=opp/hyp and cos(x)=adj/hyp. First construct your triangle with theta as your angle. You know that sin(theta)=x/3 This means that the leg of the triangle opposite theta is of length x and the hypotenuse is of length 3. You can use the Pythagorean theorem to find out what the adjacent length of triangle is and you know it is sqrt(9-x^2). Now you have constructed your triangle and you know that cos(theta)=adj/hyp=[sqrt(9-x^2)]/3 but since theta=sin^-1(x/3) then cos(theta)=cos(sin^-1(x/3))=[sqrt(9-x^2)]/3

And a little algebraic mistake was made above because you are right the expression above should read that sin (sin^-1(x/3))(cos(sin^-1(x/3))=(x/3)[sqrt(9-x^2)]/3=x[sqrt(9-x^2)]/9

 

[mrigmaiden]

It seems like many of these explanations are somewhat convoluted. Physically speaking the integrand is the equation of a semicircle, therefore using circular functions like sine and cosine is warranted. What about this? x^2 + y^2=r^2, here r=3 recall the usual parameterization of the equation in terms of sines and cosines=> x=rcos(theta) y=rsin(theta) For this case x=3cos(theta) Then use the trigonometric identity that sin^2(theta)+cos^2(theta)=1 Transform integrand from dx to dtheta and proceed. I think the solution is easier to find that way. Then in the end you can use trig inverse functions to get explicit expression for theta in terms of x.

 

[neshepard]

Wow did my review problem flow into a bad area. Sorry. As to my having pre-calc, yes. and I got a B in part 1 and 2. (As an aside, I had to teach myself part 1 during summer school because the teachers favorite expression is "I do not have to legally tell you this" as if we were in law school instead of community college. But I digress.

To tell you where my numbers came from, and please realize I could very well still be wrong, or stumbled onto the right answer blindly, or mistyped (not my best area).

Jegues, and all, you are correct in the multiplication error. I missed that on my paper and then typed it in wrong.

As to the sin(arcsin) and cos(arcsin), I know sin(arcsin(x/3))=x/3 and I cheated and looked up cos(arcsin(x))=√(1-x^2) from one of my formula memorization sheets from class. So cos(arcsin(x/3)=√(9-x^2)/3 but this is one of those I don't know why. This is why I need to get to the nitty gritty of a problem, just being told something is something doesn't help.

This whole problem is a definite integral and in working out the answer I got what my answer key told me. So somewhere in the definite integral I screwed up and lucked into the answer at the same time. lol

 

[Dickfore]

community college

Ah, that explains everything.

 

[sponsoredwalk]

Ah, that explains everything.

 

[mrigmaiden]

As to the sin(arcsin) and cos(arcsin), I know sin(arcsin(x/3))=x/3 and I cheated and looked up cos(arcsin(x))=√(1-x^2) from one of my formula memorization sheets from class. So cos(arcsin(x/3)=√(9-x^2)/3 but this is one of those I don't know why.

Consider the triangles that the other poster with the videos showed. That gives you all the information you need. Sometimes it is easy to get confused with the trigonometric inverse functions so here is an example to lubricate your brain a bit:

arcsin(1)=90 degrees=pi/2 radians That is to say that the angle that produces a sin equal to 1 is pi/2 radians. so arctrigfunction(number)=ANGLE (in degrees or radians)

Now to our problem, let's examine the expression cos(arcsin(x/3))=√(9-x^2)/3 by visualizing its associated right triangle. The triangle has theta as the angle of interest, the hypotenuse is of length 3, the leg adjacent to theta has length sqrt(9-x^2) and the leg opposite to theta has length x. Use the Pythagorean Theorem to prove that to yourself.

Now that we know what the triangle looks like, we can figure out how to evaluate the expression cos(arcsin(x/3)). Consider rephrasing the problem in your mind in the following way:

1.arcsin(x/3) gives me the ANGLE theta where sin(theta)=x/3

2.cos(arcsin(x/3)) means What is the cosine of the angle theta?

3.This verbal re-framing should help to parse the expression a bit more.

4.Recall that for right triangles:
a)cos(theta)=adjacent side length/hypotenuse length
b)sin(theta)=opposite side length/hypotenuse length
c)tan(theta)=opposite side length/adjacent side length

5.Now that we have the triangle and we have called arcsin(x/3)=theta, then cos(arcsin(x/3))=cos(theta)
=adjacent side length to theta/hypotenuse length
=(sqrt(9-x^2))/3

6. Sometimes it is better to learn how to derive expressions from first principles instead of looking them up all the time. The you can memorize the formulas after you understand how they came about. I'll explain briefly why cos(arcsin(x))=√(1-x^2) according to your class notes.

7. First draw the triangle arcsin(x)=theta=> sin(theta)=x/1=opp/hyp thus the opposite side length to theta has length equal to x and the hypotenuse has length equal to 1. Now, use the Pythagorean theorem to show that the side length adjacent to theta is of length sqrt(1-x^2). So the triangle has theta as the angle of interest, the adjacent side to theta is of length sqrt(1-x^2), the opposite side to theta is of length x and the hypotenuse is of length 1. So cos(arcsin(x))=cos(theta)=adjacent/hypotenuse=sqrt(1-x^2)/1=sqrt(1-x^2)

I hope this helps.

 

[mrigmaiden]

hehehe He's just trying to get a rise out of someone. Don't let it bother you. Take what he has to say regarding the math and physics and ignore the rest lol!