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Tamil Nadu Board of Secondary EducationHSC Science Class 12

A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored 5 km from a straight shore line. How fast is the beam moving along the shoreline when it makes an angle

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Question

A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored 5 km from a straight shoreline. How fast is the beam moving along the shoreline when it makes an angle of 45° with the shore?

Sum
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Solution


Time for one revolution = 10 sec

Now, angular velocity `"dv"/"dt" = (2pi)/10 = pi/5`

From the figure, tan 45° = `"AB"/"OA"`

1 = `x/5`

⇒ x = 5

Again, tan θ = `x/5`

x = 5 tan θ

Differentiating w.r.t. ‘t’

`("d"x)/"dt" = 5 sec^2theta  ("d"theta)/"dt"`

= `5 sec^2 (45^circ) (pi/5)`

= `(sqrt(2))^2pi`

= 2π

∴ The beam is moving at the rate of 2π km/sec.

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Chapter 7: Applications of Differential Calculus - Exercise 7.1 [Page 8]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Applications of Differential Calculus
Exercise 7.1 | Q 7 | Page 8

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