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Math Help - Problem with trigonomic derivative (pretty simple)

  1. #1
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    Problem with trigonomic derivative (pretty simple)

    I know how to derivate sinx, cosx and standard equations like

    y = x^4 + sqrt.(x)
    y' = 4x^3 + x^-1/2

    y = sinx + cosx
    y' = cosx - sinx

    and so on. But now they want me to "decide an equation for the tangent to the curve":

    a) y = cosx (pi/2, 0)

    Answer: y = -x + pi/2

    b) y = 2x - 5sinx (pi, 2pi)

    Answer: y = 7x -5pi

    I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

    Appreciate help!!
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  2. #2
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    Quote Originally Posted by vane505 View Post
    I know how to derivate sinx, cosx and standard equations like

    y = x^4 + sqrt.(x)
    y' = 4x^3 + x^-1/2

    y = sinx + cosx
    y' = cosx - sinx

    and so on. But now they want me to "decide an equation for the tangent to the curve":

    a) y = cosx (pi/2, 0)

    Answer: y = -x + pi/2

    b) y = 2x - 5sinx (pi, 2pi)

    Answer: y = 7x -5pi

    I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

    Appreciate help!!
    I take it that the numbers in the brackets are the co-ordinates of the point the tangent should touch...

    a) \frac{dy}{dx} = -\sin{x}.

    At the point x = \frac{\pi}{2}, \frac{dy}{dx} = -\sin{\frac{\pi}{2}} = -1.

    Remember that the equation for a line is given by y = mx + c where m is the slope and c is the y-intercept. Since we have the slope of the tangent line and a point that lies on the line, we can solve for c.

    y = mx + c

    0 = -1\left(\frac{\pi}{2}\right) + c

    0 = -\frac{\pi}{2} + c

    c = \frac{\pi}{2}.

    So the equation of the tangent line to the curve at the point given is

    y = -x + \frac{\pi}{2}.
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  3. #3
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    Quote Originally Posted by vane505 View Post
    I know how to derivate sinx, cosx and standard equations like

    y = x^4 + sqrt.(x)
    y' = 4x^3 + x^-1/2

    y = sinx + cosx
    y' = cosx - sinx

    and so on. But now they want me to "decide an equation for the tangent to the curve":

    a) y = cosx (pi/2, 0)

    Answer: y = -x + pi/2

    b) y = 2x - 5sinx (pi, 2pi)

    Answer: y = 7x -5pi

    I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

    Appreciate help!!
    b) y = 2x - 5\sin{x} so \frac{dy}{dx} = 2 - 5\cos{x}.

    At the point x = \pi, \frac{dy}{dx} = 2 - 5\cos{\pi} = 2 - 5(-1) = 7. This is m.

    So now we put x, y, m into y = mx + c and solve for c.

    y = mx + c

    2\pi = 7\pi + c

    c = -5\pi.


    So the equation of the tangent line at the point given is

    y = 7x - 5\pi.
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    Great! Thank you very much for your help and very clear explenation, I get it now
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