Question

A horse runs along a circle with a speed of \[20\] km/hr. A lantern is at the centre of the circle. A fence is along the tangent to the circle at the point at which the horse starts. The speed with which the shadow of the horse moves along the fence at the moment when it covers \[\dfrac{1}{8}\] of the circle in km/hr is

Answer 1

Hint: In the given question, we have to find the speed with which the shadow of the horse moves along the fence at the moment when it covers \[\dfrac{1}{8}\]of the circle in km/hr. Firstly, we have to find the angle and then the speed.

Complete step by step answer:
Given that the horse has moved an angle \[\theta \].
 The length of the shadow projection is
\[\Rightarrow L=R\tan \theta \]
 Angle when covered \[\dfrac{1}{8}\]of the circle =\[\theta \]=\[\dfrac{\pi }{4}\]
The time derivative of this position will give the speed v of the shadow.
\[\Rightarrow V=\dfrac{d}{dt}\left( R\tan \theta \right)=R{{\sec }^{2}}\theta \dfrac{d\theta }{dt}\]
Let \[\omega =\dfrac{d\theta }{dt}\] then we have that
\[\Rightarrow V=R\omega {{\sec }^{2}}\theta \]
Since the horse is running at a speed of \[20km/hr\] we get,
\[\Rightarrow R\omega =20\]
The speed with which the shadow of the horse moves along the fence at the moment when it covers \[\dfrac{1}{8}\] of the circle in km/hr is
\[\Rightarrow V=20{{\sec }^{2}}\dfrac{\pi }{4}=40km/hr\]

Therefore, the speed with which the shadow of the horse moves along the fence at the moment when it covers \[\dfrac{1}{8}\] of the circle in km/hr is \[40km/hr\]

Note: The question assumes that the shadow formed is due to the lantern, so the shadow will always be on the straight line joining the horse and the lantern. So, it being in the centre just enables you to find where the shadow is. We must not do any calculation mistakes and must know differentiation formulae very well and not do mistakes in differentiation like \[\dfrac{dy}{dx}\left( \tan x \right)=\sec x\] as the correct differentiation is \[\dfrac{dy}{dx}\left( \tan x \right)={{\sec }^{2}}x\].