Step-by-Step Solution

Topic: Rate of Change & Related Rates

Right Angled Triangle Diagram showing Pole and Car

Diagram representing the pole (height 5m), the car’s distance (x), and the angle of elevation (θ).

Given Parameters

  • • Height of pole (Opposite side) = 5 m
  • • Distance of car (Adjacent side) = x m
  • • Speed of car (dx/dt) = 20 m/s
  • • Angle of elevation = θ

Note: Since the car is moving away, x is increasing, so dx/dt is positive.

(I) RELATION BETWEEN θ AND x

In the right-angled triangle formed by the pole, the ground, and the line of sight:

$$ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{x} $$

$$ \theta = \tan^{-1}\left(\frac{5}{x}\right) $$

(II) FIND dθ/dx

Differentiate θ with respect to x using the chain rule:

$$ \frac{d\theta}{dx} = \frac{d}{dx}\left(\tan^{-1}\left(\frac{5}{x}\right)\right) $$

$$ \frac{d\theta}{dx} = \frac{1}{1 + (\frac{5}{x})^2} \times \frac{d}{dx}\left(\frac{5}{x}\right) $$

$$ \frac{d\theta}{dx} = \frac{1}{1 + \frac{25}{x^2}} \times \left(-\frac{5}{x^2}\right) $$

Simplify the fraction:

$$ \frac{d\theta}{dx} = \frac{x^2}{x^2 + 25} \times \frac{-5}{x^2} $$

$$ \frac{d\theta}{dx} = \frac{-5}{x^2 + 25} $$

(III)(A) RATE OF CHANGE (dθ/dt) AT x = 50

Using the Chain Rule:

$$ \frac{d\theta}{dt} = \frac{d\theta}{dx} \times \frac{dx}{dt} $$

Substitute the formula from (ii) and given speed v = 20:

$$ \frac{d\theta}{dt} = \left(\frac{-5}{x^2 + 25}\right) \times 20 = \frac{-100}{x^2 + 25} $$

At x = 50 m:

$$ \frac{d\theta}{dt} = \frac{-100}{(50)^2 + 25} = \frac{-100}{2500 + 25} $$

$$ \frac{d\theta}{dt} = \frac{-100}{2525} = \frac{-4}{101} \text{ rad/s} $$

Rate = -4/101 rad/s (Angle is decreasing)

(III)(B) FIND SPEED GIVEN RATE 3/101

We are given the magnitude of rate of change: |dθ/dt| = 3/101 rad/s at x = 50 m.

Let speed be v.

$$ \left| \frac{d\theta}{dt} \right| = \left| \frac{-5}{x^2 + 25} \times v \right| $$

Substitute known values (x = 50):

$$ \frac{3}{101} = \frac{5}{2525} \times v $$

$$ \frac{3}{101} = \frac{1}{505} \times v $$

Solve for v:

$$ v = \frac{3 \times 505}{101} $$

Since 505 / 101 = 5:

$$ v = 3 \times 5 = 15 \text{ m/s} $$

Speed of car = 15 m/s

⚠️ Silly Mistake Audit: The Chain Rule & Units

Don’t forget dx/dt!
A classic mistake in “Rate of Change” problems is finding the derivative with respect to position (dθ/dx) and forgetting to multiply by the velocity (dx/dt) to get the time derivative (dθ/dt). Always check if the question asks for “rate w.r.t distance” or “rate w.r.t time”.

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