(C) Given the equations $x = ct$ and $y = \frac{c}{t}$.We need to find $\frac{dy}{dt}$ at $t = 2$.Since $y = \frac{c}{t} = c \cdot t^{-1}$,we differen

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(C) Given the equations $x = ct$ and $y = \frac{c}{t}$.We need to find $\frac{dy}{dt}$ at $t = 2$.Since $y = \frac{c}{t} = c \cdot t^{-1}$,we differentiate with respect to $t$:$\frac{dy}{dt} = c \cdot (-1) \cdot t^{-2} = -\frac{c}{t^2}$.At $t = 2$,we substitute the value into the derivative:$\left. \frac{dy}{dt} \right|_{t=2} = -\frac{c}{(2)^2} = -\frac{c}{4}$.

Class 12 Mathematics Continuity and Differentiability Derivatives of Functions in Parametric Forms

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If $x = ct$ and $y = \frac{c}{t}$,find $\frac{dy}{dt}$ at $t = 2$.

KCET 2015 All KCET

A
$\frac{1}{4}$
B
$4$
C
$-\frac{1}{4}$
D
$0$

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Continuity and Differentiability

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Derivatives of Functions in Parametric Forms

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If $x = ct$ and $y = \frac{c}{t}$,find $\frac{dy}{dt}$ at $t = 2$.

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