Please help with q14 both parts thank you!

  (i) y = 5/7x

  (ii) 7/3 m/s

Step-by-step explanation:

Shadow problems almost always involve similar triangles, meaning that shadow lengths and object heights generally have the same ratio. We can use this fact to write proportions related to the measures in this problem.

The ratio of shadow-tip distance to the object height is the same for both objects. For distances in meters, we have ...

  x/7 = (x-y)/2

  2x = 7(x -y) . . . . . multiply by 14

  7y = 5x . . . . . . . add 7y-2x

  y = 5/7x . . . . . divide by the coefficient of y

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Solving the equation from the first part for x, we have ...

  x = 7/5y . . . . . multiply by 7/5

The derivative with respect to time is then ...

  x' = 7/5y'

The rate of change of y is given as 5/3 m/s, so the rate of change of x will be ...

  x' = (7/5)(5/3 m/s) = 7/3 m/s

The top of the man's shadow is moving at 7/3 meters per second.