Application of Derivatives 4 Question 56
59. A swimmer $S$ is in the sea at a distance $d km$ from the closest point $A$ on a straight shore. The house of the swimmer is on the shore at a distance $L km$ from $A$. He can swim at a speed of $u km / h$ and walk at a speed of $v km / h(v>u)$. At what point on the shore should be land so that he reaches his house in the shortest possible time?
(1983, 2M)
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Answer:
Correct Answer: 59. (1)
Solution:
- Let the house of the swimmer be at $B$.
$\therefore \quad A B=L km$
Let the swimmer land at $C$ on the shore and let
$$ A C=x km $$
$$ \begin{array}{ll} \therefore & S C=\sqrt{x^{2}+d^{2}} \text { and } \quad C B=(L-x) \\ \therefore & \text { Time }=\frac{\text { Distance }}{\text { Speed }} \end{array} $$
Time from $S$ to $B=$ Time from $S$ to $C+$ Time from $C$ to $B$
$\therefore \quad T=\frac{\sqrt{x^{2}+d^{2}}}{u}+\frac{L-x}{v}$
Let $\quad f(x)=T=\frac{1}{u} \sqrt{x^{2}+d^{2}}+\frac{L}{v}-\frac{x}{v}$
$\Rightarrow \quad f^{\prime}(x)=\frac{1}{u} \cdot \frac{1 \cdot 2 x}{2 \sqrt{x^{2}+d^{2}}}+0-\frac{1}{v}$
For maximum or minimum, put $f^{\prime}(x)=0$
$\Rightarrow \quad v^{2} x^{2}=u^{2}\left(x^{2}+d^{2}\right)$
$\Rightarrow \quad x^{2}=\frac{u^{2} d^{2}}{v^{2}-u^{2}}$
$\therefore \quad f^{\prime}(x)=0$ at $x= \pm \frac{u d}{\sqrt{v^{2}-u^{2}}},(v>u)$
But $\quad x \neq \frac{-u d}{\sqrt{v^{2}-u^{2}}}$
$\therefore$ We consider, $x=\frac{u d}{\sqrt{v^{2}-u^{2}}}$
Now, $\quad f^{\prime \prime}(x)=\frac{1}{u} \frac{d^{2}}{\sqrt{x^{2}+d^{2}}\left(x^{2}+d^{2}\right)}>0, \forall x$
Hence, $f$ has minimum at $x=\frac{u d}{\sqrt{v^{2}-u^{2}}}$.
