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Question

(A) Given the parametric equations $x = 4t$ and $y = \frac{4}{t}$.Differentiating $x$ with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}(4t) = 4$.Diffe

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$-\frac{1}{t^2}$

Vedclass · Answer

(A) Given the parametric equations $x = 4t$ and $y = \frac{4}{t}$.Differentiating $x$ with respect to $t$:$\frac{dx}{dt} = \frac{d}{dt}(4t) = 4$.Differentiating $y$ with respect to $t$:$\frac{dy}{dt} = \frac{d}{dt}\left(\frac{4}{t}\right) = 4 \cdot \frac{d}{dt}(t^{-1}) = 4 \cdot (-1 \cdot t^{-2}) = -\frac{4}{t^2}$.Using the chain rule for parametric differentiation:$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{4}{t^2}}{4} = -\frac{1}{t^2}$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=4t$ and $y=\frac{4}{t}$.

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