SATHEE: Application of Derivatives 4 Question 56

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 $B$ .

$∴ A B = L km$

Let the swimmer land at $C$ on the shore and let

$A C = x km$

$\begin{array}{ll} ∴ & S C = \sqrt{x^{2} + d^{2}} and C B = (L - x) \\ ∴ & Time = \frac{Distance}{Speed} \end{array}$

Time from $S$ to $B =$ Time from $S$ to $C +$ Time from $C$ to $B$

$∴ T = \frac{\sqrt{x^{2} + d^{2}}}{u} + \frac{L - x}{v}$

Let $f (x) = T = \frac{1}{u} \sqrt{x^{2} + d^{2}} + \frac{L}{v} - \frac{x}{v}$

$\Rightarrow f^{'} (x) = \frac{1}{u} \cdot \frac{1 \cdot 2 x}{2 \sqrt{x^{2} + d^{2}}} + 0 - \frac{1}{v}$

For maximum or minimum, put $f^{'} (x) = 0$

$\Rightarrow v^{2} x^{2} = u^{2} (x^{2} + d^{2})$

$\Rightarrow x^{2} = \frac{u^{2} d^{2}}{v^{2} - u^{2}}$

$∴ f^{'} (x) = 0$ at $x = \pm \frac{u d}{\sqrt{v^{2} - u^{2}}}, (v > u)$

But $x \neq \frac{- u d}{\sqrt{v^{2} - u^{2}}}$

$∴$ We consider, $x = \frac{u d}{\sqrt{v^{2} - u^{2}}}$

Now, $f^{''} (x) = \frac{1}{u} \frac{d^{2}}{\sqrt{x^{2} + d^{2}} (x^{2} + d^{2})} > 0, \forall x$

Hence, $f$ has minimum at $x = \frac{u d}{\sqrt{v^{2} - u^{2}}}$ .

Application of Derivatives 4 Question 56

Answer:

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Application of Derivatives 4 Question 56

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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