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Math Help - even more trig questions!

  1. #1
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    even more trig questions!

    I have another question =)

    First I'll attach the image so that it's easier to follow:
    even more trig questions!-screen-shot-2010-06-25-20.06.43.png

    For part a) I tried getting to the required answer but haven't been able to. I'll write down what I've done. Could someone tell me where I've gone wrong?

    First I isolated t
    so from x=2sint ===> t = arcsin(x/2)

    then to get a cartesian equation, I've put t back into the y equation:

    y=ln(sec(arcsin(x/2)))
    which simplifies to y=ln(2/√(4-x^2))

    differentiating i get:
    (here im not sure if its right)

    dy/dx=-x

    then when t = pi/3
    x=√3
    y=ln2

    so y - ln2=-√3 (x-√3)

    when y=0
    √3.x=3 + ln2
    so x= 1/√3 . (3 + ln2)

    but that isn't the answer....
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  2. #2
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    You're aiming at the right thing. You have the right overall view of the problem. But your derivative is wrong. Check that out.

    Another way to do this is to note that \frac{dy}{dx}=\left(\frac{dy}{dt}\right)/\left(\frac{dx}{dt}\right), and just take the derivatives of the parametric representations.
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  3. #3
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    doing it that way i get dy/dx = √3 which doesn't work either.

    Here is what I'm doing:

    x=2sint
    dx/dt = 2cost

    y=ln(sect)
    e^y=sect
    (e^y)(dy/dt)= (sect.tant)
    dy/dt= (sect.tant)/(e^y)

    ==> dy/dx = ((sect.tant)/(e^y)) x (1/2cost)

    then putting t=arcsin(x/2) and y=ln2
    i get dy/dx= √3

    But that gives me x=(1/√3)(3 + ln2)
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  4. #4
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    You've made more work for yourself than necessary. But I definitely agree with the slope of your line. What's the equation of your tangent line?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    You've made more work for yourself than necessary. But I definitely agree with the slope of your line. What's the equation of your tangent line?
    the tangent has a gradient of √3 and passes through the point (√3, ln2)

    thus its equation is

    y-ln2 = √3(x-√3)
    y=(√3)x - 3 + ln2

    at point A, y = 0

    thus (√3)x - 3 + ln2 = 0
    which suggests that x= (1/√3)(3 - ln2)
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  6. #6
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    Well, isn't your answer correct, now? You had a sign error in post #3 in conjunction with the \ln(2), but that appears to be fixed now. Also, note that \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}. So, aren't you done with this part?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    Well, isn't your answer correct, now? You had a sign error in post #3 in conjunction with the \ln(2), but that appears to be fixed now. Also, note that \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}. So, aren't you done with this part?
    how can i know in an exam that \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} without a calculator? I just don't see it for some reason..

    I'll have a go at part b) now =)
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  8. #8
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    Multiply the top and bottom of the LHS by \sqrt{3}.
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  9. #9
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    haha, please forget i ever asked that!

    For the second part my intuition is to integrate the curve's cartesian equation y= ln(2/√(4-x^2)) between (√3/3)(3-ln2) and 0

    and then integrate the curve again between √3 and y= ln(2/√(4-x^2))

    add those two, and then subtract the area of the right angle triangle formed between x=A x=√3 and P.

    If this is a good way of doing it, when integrating ln(2/√(4-x^2))
    do you have to use integration by parts?

    If so, if you let u=ln(2/√(4-x^2)) then du/dx here will be the same integral as before but this time replace the t's with t=arcsin(x/2) and the y by y=ln(sect)?

    Thank you
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  10. #10
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    Hey, we all have brain freezes.

    Again, I think you're making things more difficult than you need. Just integrate the curve from 0 to \sqrt{3}, and then subtract the area of the triangle. There's no need to break up the integral of the upper curve into two pieces.

    As for finding the antiderivative of the function, I think I would probably go with a trig substitution first. What do you think?
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    Hey, we all have brain freezes.

    Again, I think you're making things more difficult than you need. Just integrate the curve from 0 to \sqrt{3}, and then subtract the area of the triangle. There's no need to break up the integral of the upper curve into two pieces.

    As for finding the antiderivative of the function, I think I would probably go with a trig substitution first. What do you think?
    no integration by parts?
    It's be easy if it were just y= 2/√(4-x^2) but the ln of y= ln(2/√(4-x^2)) throws me off..

    i'm not sure what trig substitution to use..
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  12. #12
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    I do trig substitutions all the same way.

    1. I draw a right triangle and label the right angle.
    2. Look at the signs of the two squared terms. If they have the same sign, then let each side be an unsquared term, and compute the hypotenuse using Pythag. If they have the opposite sign, let the positive unsquared term be the hypotenuse, and the negative unsquared term be one of the sides. Compute the remaining side using Pythag. Label all sides.
    3. Assign the angle theta and label on drawing.
    4. Transfer the integrand, the differential, and the limits (if you're doing a definite integral) into the theta domain.
    5. Perform the rest of the integration in the theta domain. If you have a definite integral, you can finish here. Otherwise,
    6. In the case of an indefinite integral, transfer back to the original variable's domain. You can use your triangle that you drew and labeled in order to do this easily.

    So, what do you think now? Where would you start?
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  13. #13
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    Quote Originally Posted by Ackbeet View Post
    I do trig substitutions all the same way.


    4. Transfer the integrand, the differential, and the limits (if you're doing a definite integral) into the theta domain.
    5. Perform the rest of the integration in the theta domain. If you have a definite integral, you can finish here. Otherwise,
    6. In the case of an indefinite integral, transfer back to the original variable's domain. You can use your triangle that you drew and labeled in order to do this easily.

    So, what do you think now? Where would you start?
    I don't really understand what you mean in number 4..
    I've drawn the triangle and labeled all the sides and theta. I have my integrand y= ln(2/√(4-x^2)), what do you mean by differential? what I get when I differentiate y= ln(2/√(4-x^2))? and then I have my limits 0 and √3.

    what do you mean by transfer these into the theta domain?
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  14. #14
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    Do you mean for me to get to:

    dy/dx = x / (4-x^2)
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  15. #15
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    So, number 4 works like this. You've got your integral,

    \int_{0}^{\sqrt{3}}\ln\left(\frac{2}{\sqrt{4-x^{2}}}\right)dx.

    Now then, suppose you draw your triangle with theta as the lower left angle. The squared terms have opposite signs, so you let 2 be the hypotenuse, and x the opposite side to theta. That means the adjacent side has length \sqrt{4-x^{2}}. You need to transform the limits, the integrand, and the differential so that there are no more x's, only theta's. Here are some relevant equations that let you do that:

    \frac{x}{2}=\sin(\theta)

    \frac{2}{\sqrt{4-x^{2}}}=\sec(\theta).

    So, hopefully, getting the integrand to have only theta's should be fairly straight-forward. How do you propose to get the limits and the differential to have only theta's?
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