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Math Help - Differentials and apporximations help ple

  1. #1
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    Differentials and apporximations help ple

    Usually I get that one problem a night that I just get completely locked up on and was hoping for an explanation on where I am going wrong. As always thanks for your help, ya'll have been a life saver. Problem below

    s= Sqrt. (t^2 - cot*t + 2)^3 Find ds
    so I am looking for the differential right? Since its a square root problem I believe I sould be taking diff and putting it in s'= Sqrt[(t^2- cot*t + 2)^3] '
    That would go to Sqrt [(t^2 - cot*t + 2)^3'] * dt. Then my algerbra is kicking my butt. Why cant I pull out a (t^2-cot*t + 2)'Sqrt(t^2-cot*t+2)'(dt). then take thir deriv. This stuff kills me,
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    Re: Differentials and apporximations help ple

    Quote Originally Posted by psilver1 View Post
    Usually I get that one problem a night that I just get completely locked up on and was hoping for an explanation on where I am going wrong. As always thanks for your help, ya'll have been a life saver. Problem below

    s= Sqrt. (t^2 - cot*t + 2)^3 Find ds
    so I am looking for the differential right? Since its a square root problem I believe I sould be taking diff and putting it in s'= Sqrt[(t^2- cot*t + 2)^3] '
    That would go to Sqrt [(t^2 - cot*t + 2)^3'] * dt. Then my algerbra is kicking my butt. Why cant I pull out a (t^2-cot*t + 2)'Sqrt(t^2-cot*t+2)'(dt). then take thir deriv. This stuff kills me,
    If you're trying to find the derivative, you need to use the Chain Rule. \displaystyle \begin{align*} \frac{ds}{dt} = \frac{ds}{du} \cdot \frac{du}{dt} \end{align*}.

    Your function is \displaystyle \begin{align*} s= \sqrt{\left[t^2 - \cot{(t)} + 2\right]^3} = \left[ t^2 - \cot{(t)} + 2 \right]^{\frac{3}{2}} \end{align*}, so let \displaystyle \begin{align*} u = t^2 - \cot{(t)} + 2 \implies s = u^{\frac{3}{2}} \end{align*}. You may need to use the fact that \displaystyle \begin{align*} \cot{(t)} = \frac{\cos{(t)}}{\sin{(t)}} \end{align*} and the quotient rule to evaluate the derivative of \displaystyle \begin{align*} \cot{(t)} \end{align*}.
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    Re: Differentials and apporximations help ple

    We just learned this material today, last week we did stuff like dy/dx and d/dx and this says just use dy. Isn't that different than derivatives. I have been wrong a lot but this says differentials. So find ds. s=Sqrt(t^2 - cot(t) + 2)(dx) and then break it down?? I dont understand what I should do to break it down. If I use the chain rule does the Sqrt act as the outside and the (T^2 - cot(t) + 2)(dx) act as the inside?
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    Re: Differentials and apporximations help ple

    Hello, psilver11

    s \:=\:\sqrt{(t^2 - \cot t + 2)^3}\;\;\;\text{Find }ds.

    We have: . s \;=\;(t^2-\cot t + 2)^{\frac{3}{2}}

    Then: . ds \;=\;\tfrac{3}{2}(t^2-\cot t+ 2)^{\frac{1}{2}}(2t + \csc^2t)\,dt

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    Re: Differentials and apporximations help ple

    If you dont mind me asking where does the csc^2t come into play. I understand the 3/2(t^2 - cot t + 2)^1/2 using the product rule and. Also this is Algerbra but to pul it out of the sqrt you raised it to 3/2, well it was raised to 3 but then you raised it by 2? in order to get the 3/2?
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    Re: Differentials and apporximations help ple

    Quote Originally Posted by psilver1 View Post
    If you dont mind me asking where does the csc^2t come into play. I understand the 3/2(t^2 - cot t + 2)^1/2 using the product rule and. Also this is Algerbra but to pul it out of the sqrt you raised it to 3/2, well it was raised to 3 but then you raised it by 2? in order to get the 3/2?
    It's NOT using the Product Rule, it's the Chain Rule. And like I told you, use the Quotient Rule to find the derivative of cot(t).

    Even though it's not technically a fraction, \displaystyle \begin{align*} \frac{dy}{dx} = f(x) \end{align*} works similarly to a fraction, so \displaystyle \begin{align*} dy = f(x)\,dx \end{align*}
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