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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 6 · Set 1 · Q37
A ladder of fixed length ' h ' is to be placed along the wall such that it is free to move along the height of the wall. Based upon the above…
A ladder of fixed length ' $\displaystyle \mathrm{h}$ ' is to be placed along the wall such that it is free to move along the height of the wall. Based upon the above information, answer the following questions :
(i)
Express the distance ( y ) between the wall and foot of the ladder in terms of ' $\displaystyle \mathrm{h}$ ' and height ( $\displaystyle \mathrm{x}$ ) on the wall at a certain instant. Also, write an expression in terms of $\displaystyle \mathrm{h}$ and $\displaystyle \mathrm{x}$ for the area (A) of the right triangle, as seen from the side by an observer.
(ii)
Find the derivative of the area (A) with respect to the height on the wall ( x ), and find its critical point.
(iii)
(a)
Show that the area (A) of the right triangle is maximum at the critical point.
OR
(b)
If the foot of the ladder whose length is $\displaystyle 5$ m , is being pulled towards the wall such that the rate of decrease of distance
(y) is $\displaystyle 2 \mathrm{~m} / \mathrm{s}$, then at what rate is the height on the wall
(x)
increasing, when the foot of the ladder is $\displaystyle 3$ m away from the wall? Case Study - $\displaystyle 3$
Marking-scheme solution
(i)
\(\displaystyle \mathrm{y}^{2}=\mathrm{h}^{2}-\mathrm{x}^{2}\)
\(\displaystyle \mathrm{A}=\frac{1}{2} \mathrm{xy}=\frac{1}{2} \mathrm{x} \sqrt{\mathrm{h}^{2}-\mathrm{x}^{2}}\)
(ii)
\(\displaystyle \frac{\mathrm{dA}}{\mathrm{dx}}=\frac{1}{2} \sqrt{\mathrm{~h}^{2}-\mathrm{x}^{2}}+\frac{1}{2} \mathrm{x} \frac{-\mathrm{x}}{\sqrt{\mathrm{h}^{2}-\mathrm{x}^{2}}}\)
\(\displaystyle \frac{\mathrm{dA}}{\mathrm{dx}}=0\) gives \(\displaystyle \mathrm{x}=\frac{\mathrm{h}}{\sqrt{2}}\)
(iii)
\(\displaystyle \mathrm{A}^{\prime \prime}=\frac{1}{2} \frac{-4 \mathrm{x} \cdot \sqrt{\mathrm{h}^{2}-\mathrm{x}^{2}}-\left(\mathrm{h}^{2}-2 \mathrm{x}^{2}\right) \dfrac{-\mathrm{x}}{\sqrt{\mathrm{h}^{2}-\mathrm{x}^{2}}}}{\mathrm{~h}^{2}-\mathrm{x}^{2}}\) is \(\displaystyle <0\) at \(\displaystyle \mathrm{x}=\frac{\mathrm{h}}{\sqrt{2}}\)
Hence A is maximum at critical point
\(\displaystyle \mathrm{y}^{2}=25-\mathrm{x}^{2}\) hence \(\displaystyle \mathrm{y}=3\) gives \(\displaystyle \mathrm{x}=4\)
\(\displaystyle 2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dt}}=-2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}\)
\(\displaystyle \frac{\mathrm{dx}}{\mathrm{dt}}=1.5 \mathrm{~m} / \mathrm{s}\)
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