A point is in motion along a hyperbola y = fr | vedclass

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(C) Given the equation of the hyperbola is $y = \frac{10}{x}$.The rate of change of the abscissa is given as $\frac{dx}{dt} = 1 	ext{ unit/s}$.To fin

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decreases at the rate of $\frac{2}{5} 	ext{ unit/s}$

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(C) Given the equation of the hyperbola is $y = \frac{10}{x}$.The rate of change of the abscissa is given as $\frac{dx}{dt} = 1 	ext{ unit/s}$.To find the rate of change of the ordinate $\frac{dy}{dt}$, we differentiate $y$ with respect to $t$ using the chain rule:$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{10}{x} ight) = -\frac{10}{x^2} \cdot \frac{dx}{dt}$.Substitute the given values $x = 5$ and $\frac{dx}{dt} = 1$ into the derivative:$\frac{dy}{dt} = -\frac{10}{(5)^2} \cdot (1) = -\frac{10}{25} = -\frac{2}{5} 	ext{ unit/s}$.Since the result is negative, the ordinate $y$ decreases at the rate of $\frac{2}{5} 	ext{ unit/s}$.

$A$ point is in motion along a hyperbola $y = \frac{10}{x}$ such that its abscissa $x$ increases uniformly at a rate of $1 \text{ unit/s}$. Find the rate of change of its ordinate when the point passes through $(5, 2)$.

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