A 10 , cm long rod AB moves with its ends on | vedclass

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Question

(A) Let $OA = x$ and $OB = y$. Since the rod $AB$ has a constant length of $10 \, cm$,by the Pythagorean theorem,we have $x^2 + y^2 = 10^2 = 100$. Dif

Vedclass · Accepted Answer

$\frac{8}{3} \, cm/sec$

Vedclass · Answer

(A) Let $OA = x$ and $OB = y$. Since the rod $AB$ has a constant length of $10 \, cm$,by the Pythagorean theorem,we have $x^2 + y^2 = 10^2 = 100$. Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$. Given that $\frac{dx}{dt} = 2 \, cm/sec$ (as $A$ moves away from $O$,$x$ increases) and at the instant $x = 8 \, cm$,we find $y$ using $x^2 + y^2 = 100$: $8^2 + y^2 = 100 \Rightarrow 64 + y^2 = 100 \Rightarrow y^2 = 36 \Rightarrow y = 6 \, cm$. Substituting these values into the differentiated equation: $8(2) + 6 \frac{dy}{dt} = 0 \Rightarrow 16 + 6 \frac{dy}{dt} = 0 \Rightarrow 6 \frac{dy}{dt} = -16 \Rightarrow \frac{dy}{dt} = -\frac{16}{6} = -\frac{8}{3} \, cm/sec$. The negative sign indicates that the distance $y$ is decreasing,meaning the end $B$ is moving towards $O$ at a rate of $\frac{8}{3} \, cm/sec$.

$A$ $10 \, cm$ long rod $AB$ moves with its ends on two mutually perpendicular straight lines $OX$ and $OY$. If the end $A$ is moving at the rate of $2 \, cm/sec$,then when the distance of $A$ from $O$ is $8 \, cm$,the rate at which the end $B$ is moving is:

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