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Math Help - How do you work out the exact gradient when y = 3cos2x?

  1. #1
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    How do you work out the exact gradient when y = 3cos2x?

    Hi, I Know what to do, find the first derivative. But i get stuck on a line which can't find the exact value.

    Question: Find the gradient of the tangent to the curve y = 3cos2x at the point where x = pi/6.

    dy/dx = -6sin2x

    y'(pi/6) = -6sin(2*pi/6)

    =-6sin(2pi/6)

    how do I simplify that? I'm totally lost as of what to do next.



    PS: On another note what does d/dx mean as compared to dy/dx?

    Thanks,

    MBB
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by MathsBeforeBedtime View Post
    Hi, I Know what to do, find the first derivative. But i get stuck on a line which can't find the exact value.

    Question: Find the gradient of the tangent to the curve y = 3cos2x at the point where x = pi/6.

    dy/dx = -6sin2x

    y'(pi/6) = -6sin(2*pi/6)

    =-6sin(2pi/6)

    how do I simplify that? I'm totally lost as of what to do next.



    PS: On another note what does d/dx mean as compared to dy/dx?

    Thanks,

    MBB
    You may need to review your unit circle to find \sin(\pi/3)...

    Also, \frac{\,d}{\,dx}\left[\cdots\right] is a way of saying that you're differentiating something. So for example \frac{\,d}{\,dx}\left[x^2+2x+1\right] means that you should differentiate x^2+2x+1 with respect to x.
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    You may need to review your unit circle to find \sin(\pi/3)...

    Also, \frac{\,d}{\,dx}\left[\cdots\right] is a way of saying that you're differentiating something. So for example \frac{\,d}{\,dx}\left[x^2+2x+1\right] means that you should differentiate x^2+2x+1 with respect to x.
    why pi/3
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  4. #4
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    Because \frac{2}{6}= \frac{1}{3}. If you couldn't see that, no wonder you have problems with this!

    Also \frac{\pi}{3} radians is the same as \frac{180}{3}= 60 degrees. And 60 degrees is the measure of each of the angles in an equilateral triangle. Imagine an equilateral triangle with each side of length 1. Drop a perpendicular from one vertex to the opposite side. That will also bisect the opposite side so you now have two right triangles with one leg of length 1/2 and hypotenuse of length 1. By the Pythagorean theorem, 1^2= (1/2)^2+ b^2 where b is the length of the other leg- the perpendicular. Then b^2= 1- 1/4= 3/4 so b= \sqrt{3}//2. Sine is "opposite side over hypotenuse so sin(\pi/3)= sin(60)= \sqrt{3}/2.
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