Related Rates of Change

Related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another, using the chain rule to link them. These problems appear frequently in Further Mathematics exams and require careful setup.

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The Chain Rule for Related Rates

If $y$y depends on $x$x, and $x$x depends on $t$t (time), then:

$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$dtdy​=dxdy​⋅dtdx​

More generally, for three or more linked variables:

$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$dtdA​=drdA​⋅dtdr​

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Strategy for Solving Related Rates Problems

1. Identify all the quantities and which rates are given/required.
2. Find a relationship (formula) linking the quantities.
3. Differentiate the relationship with respect to time $t$t.
4. Substitute the known values and solve for the unknown rate.

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Worked Examples

Example 1: A circle has a radius that is increasing at a rate of 2 cm/s. Find the rate of increase of the area when the radius is 5 cm.

Given: $\frac{dr}{dt} = 2$dtdr​=2. Find: $\frac{dA}{dt}$dtdA​ when $r = 5$r=5.

$A = \pi r^2$A=πr2, so $\frac{dA}{dr} = 2\pi r$drdA​=2πr.

$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi(5)(2) = 20\pi \text{ cm}^2\text{/s}$dtdA​=drdA​⋅dtdr​=2π(5)(2)=20π cm2/s

Example 2: A spherical balloon is being inflated so that its volume increases at 100 cm³/s. Find the rate of increase of the radius when the radius is 10 cm.

Given: $\frac{dV}{dt} = 100$dtdV​=100. Find: $\frac{dr}{dt}$dtdr​ when $r = 10$r=10.

$V = \frac{4}{3}\pi r^3$V=34​πr3, so $\frac{dV}{dr} = 4\pi r^2$drdV​=4πr2.

$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$dtdV​=4πr2⋅dtdr​

$100 = 4\pi(100)\frac{dr}{dt}$100=4π(100)dtdr​

$\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.0796 \text{ cm/s}$dtdr​=400π100​=4π1​≈0.0796 cm/s

Example 3: A ladder 5 m long leans against a vertical wall. The foot of the ladder slides away from the wall at 0.5 m/s. Find the rate at which the top of the ladder slides down the wall when the foot is 3 m from the wall.

Let $x$x = distance of foot from wall, $y$y = height of top.

$x^2 + y^2 = 25$x2+y2=25. Differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$2xdtdx​+2ydtdy​=0.

When $x = 3$x=3: $y = 4$y=4. Given $\frac{dx}{dt} = 0.5$dtdx​=0.5:

$2(3)(0.5) + 2(4)\frac{dy}{dt} = 0$2(3)(0.5)+2(4)dtdy​=0 $3 + 8\frac{dy}{dt} = 0$3+8dtdy​=0 $\frac{dy}{dt} = -\frac{3}{8} \text{ m/s}$dtdy​=−83​ m/s

The top slides down at $\frac{3}{8}$83​ m/s (the negative sign indicates $y$y is decreasing).

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Connected Variables