Mathematics: Me differentiate this a curve is defined by the parametric equations x=t^2 and y=t^3 show that the equation of the tangent to the curve at the point p (p^2, p^3) is 2y-3px+p^3=0

Me differentiate this a curve is defined by the parametric

Ответ:

ptanner706

11.03.2022

Hello,

$x=t^2===\ \textgreater \ dx=2tdt\\\\
y=t^3===\ \textgreater \ dy=3t^2dt\\\\
 \dfrac{dy}{dx} = \dfrac{3}{2} t\\\\
P=(p^2;p^3)===\ \textgreater \ t=p\\\\
Equation\ of\ the\ tangent:\\
y-p^3= \frac{3}{2}p(x-p^2)\\
===\ \textgreater \ 2y-3px+p^3=0\\\\
$

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Ответ:

mgarcia1394

28.11.2021

Mathematics

The first question is 83,63. Tue second is 75, and the third is the sun,the pen , and the ice cream

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Me differentiate this a curve is defined by the parametric equations x=t^2 and y=t^3 show that the equation of the tangent to the curve at the point p (p^2, p^3) is 2y-3px+p^3=0

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