The ordinate of a point describing the circle | vedclass

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Question

(C) Given the equation of the circle is $x^2 + y^2 = 25$. Differentiating both sides with respect to time $t$,we get: $2x \frac{dx}{dt} + 2y \frac{dy}

Vedclass · Accepted Answer

$\frac{3}{4} \, cm/sec$

Vedclass · Answer

(C) Given the equation of the circle is $x^2 + y^2 = 25$. Differentiating both sides with respect to time $t$,we get: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ $\Rightarrow x \frac{dx}{dt} + y \frac{dy}{dt} = 0$. Given that the ordinate $y$ decreases at the rate of $1 \, cm/sec$,so $\frac{dy}{dt} = -1 \, cm/sec$. When $y = 3 \, cm$,we find $x$ using the circle equation: $x^2 + 3^2 = 25 \Rightarrow x^2 = 25 - 9 = 16 \Rightarrow x = 4 \, cm$ (since $x > 0$). Substituting the values into the derivative equation: $4 \frac{dx}{dt} + 3(-1) = 0$ $4 \frac{dx}{dt} = 3$ $\frac{dx}{dt} = \frac{3}{4} \, cm/sec$. Thus,the abscissa increases at the rate of $\frac{3}{4} \, cm/sec$.

The ordinate of a point describing the circle $x^2 + y^2 = 25$ decreases at the rate of $1 \, cm/sec$. Find the rate of change of the abscissa of the point when the ordinate is equal to $3 \, cm$ (Given $x > 0, y > 0$).

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