Question

x²= at²+bt³

y³= at⁴+bt⁸
How will we differentiate both sides with respect to t (two seperate equation)​

Answer 1

Answer:

$\longrightarrow\sf{\dfrac{dx}{dt} = \dfrac{2at+3bt^{2}}{2x}}$

$\longrightarrow\sf{\dfrac{dy}{dt} = \dfrac{4at^{3} +8bt^{7}}{3y^{2} }}$

$\longrightarrow\sf{\dfrac{dx}{dy} = \dfrac{8at^{3}x+16bt^{7}x}{6aty^{2}+9bt^{2}y^{2}}}$

Explanation:

First equation given : $\sf{x^{2} =at^{2} +bt^{2}. }$

Hence, $\sf{\dfrac{d(x^{2})}{dt}=\dfrac{d(at^{2}+bt^{3})}{dt}}$

$\leadsto \sf{2x\times\dfrac{dx}{dt}=2at+3bt^{2} }$

$\longrightarrow\bf{\dfrac{dx}{dt} = \dfrac{2at+3bt^{2}}{2x}}.....(1)$

Second equation given : $\sf{y^{3} =at^{4} +bt^{8}. }$

Hence,$\sf{\dfrac{d(y^{3})}{dt}=\dfrac{d(at^{4}+bt^{8})}{dt}}$

$\leadsto \sf{3y^{2} \times\dfrac{dy}{dt}=4at^{3} +8bt^{7} }$

$\longrightarrow\bf{\dfrac{dy}{dt} = \dfrac{4at^{3} +8bt^{7}}{3y^{2}}}.....(2)$

Now, Dividing (2) by (1).

$\longrightarrow\sf{\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} =\dfrac{dy}{dt}\times\dfrac{dt}{dx} = \dfrac{dy}{dx}}$

Hence,

$\leadsto\sf{\dfrac{4at^{2}+8bt^{7}}{3y^{2}} \times \dfrac{2x}{2at+3bt^{2}}}$

$\longrightarrow\red{\tt{\dfrac{dy}{dx} = \dfrac{8at^{3}x+16bt^{7}x}{6aty^{2}+9bt^{2}y^{2}}}}$

Answer 2

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• Refer the attachment