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Anthonyn2121

  • one year ago

Find dy/ds if y= s sqrt(1-s^2)+arccos(s)

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  1. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle y=s\sqrt{1-s^2}+\cos^{-1}s }\) What problem do you have with differentiating the above?

  2. SolomonZelman
    • one year ago
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    Can you use the *Product Rule* for the following? \(\large\color{black}{ \displaystyle s\sqrt{1-s^2} }\)

  3. Anthonyn2121
    • one year ago
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    I used the product rule and derived the arccos and got |dw:1443655706621:dw|

  4. SolomonZelman
    • one year ago
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    you forgot the chain rule for \(\sqrt{1-s^2}\)

  5. SolomonZelman
    • one year ago
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    |dw:1443655787406:dw|

  6. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle y=s\sqrt{1-s^2}+\cos^{-1}s }\) so the first part becomes, \(\large\color{black}{\dfrac{-s^2}{\sqrt{1-s^2}}+\sqrt{1-s^2}}\)

  7. Anthonyn2121
    • one year ago
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    Thanks! I know the right answer is -2s^2/sqrt(1-s^2). Will I get the right answer if I multiply the -2s I was missing?

  8. SolomonZelman
    • one year ago
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    that is not right

  9. SolomonZelman
    • one year ago
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    you have 2 on the bottom and that cancels two on top and bottom of the first term. (this 2 on the bottom comes from the power rule of the square root)

  10. SolomonZelman
    • one year ago
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    Again the first term is, \(\large\color{black}{\dfrac{-s^2}{\sqrt{1-s^2}}}\)

  11. SolomonZelman
    • one year ago
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    then, the inverse cosine is as follows: \(\large\color{black}{ \displaystyle y=\cos^{-1}s }\) \(\large\color{black}{ \displaystyle s=\cos y }\) \(\large\color{black}{ \displaystyle 1=y'(-\sin y) }\) \(\large\color{black}{ \displaystyle -1/\sqrt{1-\cos^2y}=y' }\) \(\large\color{black}{ \displaystyle -1/\sqrt{1-s^2}=y' }\)

  12. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \color{blue}{y}=\color{green}{s\sqrt{1-s^2}}+\color{red}{\cos^{-1}s} }\) \(\large\color{black}{ \displaystyle \color{blue}{dy/ds}=\color{green}{\frac{-s^2}{\sqrt{1-s^2}}+\sqrt{1-s^2}}\color{red}{-\frac{1}{\sqrt{1-s^2}}} }\)

  13. SolomonZelman
    • one year ago
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    questions?

  14. Anthonyn2121
    • one year ago
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    Nope! I got it. Thank you so much.

  15. SolomonZelman
    • one year ago
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    yw

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