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nickymarden

  • 3 years ago

Determine the tangent line to he curve y^2(2-x)=x^3 at the point P(1,1).

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  1. myko
    • 3 years ago
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    put it like this: \[y=\pm \sqrt{x ^{3}/(2-x)}\] abd find the derivative

  2. nickymarden
    • 3 years ago
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    i did \[f(1) =1\] \[(f(x))^2.(2-x)=x^3\] and derived it

  3. myininaya
    • 3 years ago
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    \[y^2(2-x)=x^3\] \[(y^2)'(2-x)+y^2(2-x)'=(x^3)' \text{ product rule }\] \[2yy'(2-x)+y^2(0-1)=3x^2\]

  4. nickymarden
    • 3 years ago
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    yeah, i did that :)

  5. myininaya
    • 3 years ago
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    \[2yy'(2-x)+y^2(-1)=3x^2\] \[2yy'(2-x)-y^2=3x^2\]

  6. myininaya
    • 3 years ago
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    \[\text{ add } y^2 \text{ on both sides }\] \[\text{ divide both sides by } 2y(2-x) \]

  7. myininaya
    • 3 years ago
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    The objective is to solve for y' Following these steps will do that y' is the slope by the way

  8. nickymarden
    • 3 years ago
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    thank you :))

  9. robtobey
    • 3 years ago
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    \[(2-x) y^2=x^3 \]Calculate the total derivative of the above\[2 (2-x) y dy-y^2 dx=3x^2 dx \]Solve for dy\[dy=-\frac{\left(3 x^2+y^2\right) dx}{2 (x-2) y} \]then divide both sides by dx\[\frac{dy}{ dx}=-\frac{\left(3 x^2+y^2\right)}{2 (x-2) y} \]Evaluate the RHS of the above at point (1,1)\[-\frac{\left(3\ 1^2+1^2\right)}{2 (1-2) 1}=2 \]

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