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Thread: Looking for help with my calculus exam practice question

  1. #1
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    Looking for help with my calculus exam practice question

    Consider f(t) = -(2t + 3) sin( t )+ 5 cos(t)

    find constants C1, C2, C3 so that the derivative of g(t) = (C1t + C2) cos(t), + C3 sin(t) is equal to f(t)



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  2. #2
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    Re: Looking for help with my calculus exam practice question

    Did you differentiate $\displaystyle (C_1t+ C_2)\cos(t)+ C_3\sin(t)$? The derivative of $\displaystyle C_1t+ C_2$ is $\displaystyle C_1$, the derivative of $\displaystyle \cos(t)$ is $\displaystyle -\sin(t)$, and the derivative of $\displaystyle \sin(t)$ is $\displaystyle cos(t)$. Of course you will need to use the "product rule" to differentiate $\displaystyle (C_1t+ C_2)\cos(t)$.
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  3. #3
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    Re: Looking for help with my calculus exam practice question

    Thanks for you response. I think I understand how to differentiate it.

    So is my C1 = 2, C2 = 3 and C3 = 5?
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  4. #4
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    Re: Looking for help with my calculus exam practice question

    Quote Originally Posted by khoile View Post
    Thanks for you response. I think I understand how to differentiate it.

    So is my C1 = 2, C2 = 3 and C3 = 5?
    No, you have the wrong $C_3$. My recommendation is exactly the same as that of HallsofIvy. I recommend differentiating $g(t)$ and see what you get. Once you do, equate coefficients.
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  5. #5
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    Re: Looking for help with my calculus exam practice question

    What did you get for the derivative?
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  6. #6
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    Re: Looking for help with my calculus exam practice question

    Not 100% sure its correct but this is what I got for the derivative:

    g’(t) = (c1t + c2) -sin(t) + cos(t)C1 c3 cos(t)
    = -(c1t + c2) sin(t) + (c1 + c3) cost (t)
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  7. #7
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    Re: Looking for help with my calculus exam practice question

    Quote Originally Posted by khoile View Post
    Not 100% sure its correct but this is what I got for the derivative:

    g(t) = (c1t + c2) -sin(t) + cos(t)C1 c3 cos(t)
    = -(c1t + c2)[sin(t)] + (c1 + c3) cost (t)
    Your derivative is correct.
    Look at this:
    $ \begin{align*}g(x)&=(C_1t+C_2)\cos(t)+C_3\sin(t) \\g'(t)&=C_1\cos(t)-(C_1+C_2)\sin(t)+C_3cos(t)\\&=-(C_1+C_2)\sin(t)+(C_1+C_3)\cos(t) \end{align*}$

    So $C_1=2, ~C_2=3~\&~C_3=~?$
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