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March 28, 2017

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Find the equation (in terms of x and y) of the tangent line to the curve r=3sin3(theta) at theta=pi/4.

I don't know if I'm taking the right approach for this question but her goes what I did:

I found the slope

I got (9(-1/2)(sqrt3/2)+3(sqrt3/2)(1/2))/(9(-1/2)(1/2)-3(sqrt3/2)(sqrt3/2))

Is that right? what else do I have to do?

  • PLEASE I really need help!!! - ,

    I don't know what to do for this problem.

  • Calculus - ,

    Since you want the equation in terms of x and y, let's change what looks a polar coordinate equation to the x-y form

    r = 3sin(3Ø)
    let k=3Ø
    so r = 3sin k
    r = 3(y/r)
    r^2 = 3y
    x^2 + y^2 = 3y
    then
    2x + 2y dy/dx = 3dy/dx
    2x/(3-2y) = dy/dx

    when Ø = π/4
    k = 3(π/4) = 3π/4
    so r = 3sin(3π/4) = 3√2/2
    but r^2 = 3y
    9/2 = 3y
    y = 3/2
    sub that into x^2 + y^2 = 3y
    x^2 + 9/4 = 9/2
    x^2 = 9/4
    x = 3/2
    so the point of contact is (3/2 , 3/2)

    dy/dx = 2(3/2) / (3-2(3/2) = 3/(0) ----> undefined !
    aaahhhh!, so the tangent line is a vertical line

    so its equation is simply x = 3/2

    check my arithmetic.

  • Calculus - ,

    Webwork asks me to enter an answer for y=

  • Calculus - ,

    A vertical line cannot be written in the form y = ...

    Will it let you enter x + 0y = 3/2 ??

    Did you look at my solution and follow the steps?
    Do you see any errors?

  • Calculus - ,

    I don't see any error, but webwork just says y= , I can't enter anything besides what y is supposed to be. The slope I took was a method I found in my book but it only says that, how to take the slope, and I don't even know if I had that right. But I cant enter anything besides the answer for y=.

  • Calculus - ,

    Leaving things in polar coordinates is a lot simpler here.

    if r = f(t)

    dy/dx = [f'(t)*sin(t) + f(t)*cos(t)] / [f'(t)*cos(t) - f(t)*sin(t)]

    t = π/4

    f = 3sin(3t) = 3/√2
    f' = 9cos(3t) = -9/√2
    sin(t) = 1/√2
    cos(t) = 1/√2

    dy/dx = (-9/√2 * 1/√2 + 3/√2 * 1/√2)/(-9/√2 * 1/√2 - 3/√2 * 1/√2)
    = (-6/2 +3/2)/(-9/2 - 3/2) = -6/-12 = 1/2

    google polar coordinates graphing and the first hit is a nice simple grapher for polar coordinates. Looks like slope = 1/2

    So, now we have a point (π/4,3/√2) and a slope: 1/2

    y-3/√2 = 1/2(x-π/4)

  • Calculus - ,

    I am not sure I'm right, but it seems to me that:
    r=3sin3(theta)
    is meant to be interpreted as
    r(t)=3sin³(θ),
    in which case dy/dx=2.
    I do not know why, but when I started looking at it, I did the same interpretation of 3θ.

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