DIFFERENTIATION · HL
Rates of change
How fast one quantity changes as another changes.
Section 1 of 6
Rates of change
$r = 1 \qquad A = \pi$ $\times 4$
$r = 2 \qquad A = 4\pi$ $\times 2\tfrac{1}{4}$
$r = 3 \qquad A = 9\pi$ $\times 1.7$
$r = 4 \qquad A = 16\pi$
Simplify
(i) $\dfrac{1}{2} \times \dfrac{2}{5} = \dfrac{1}{2} \times \dfrac{2}{5} = \dfrac{1}{5}$
(ii) $\dfrac{1}{2} \div \dfrac{2}{5} = \dfrac{1}{2} \times \dfrac{5}{2}$
Tidy if you can.
(i) $\dfrac{dA}{dt} \div \dfrac{dr}{dt} = \dfrac{dA}{dt} \times \dfrac{dt}{dr} = \dfrac{dA}{dr}$
(ii) $\dfrac{dr}{dt} \times \dfrac{dA}{dr} = \dfrac{dA}{dt}$
$d \;:\; \Delta \;:\;$ delta
Section 2 of 6
Circles
A circle is increasing at rate of $5\,\text{m/s}$ find rate of increase of area when $r = 8\,\text{m}$.
Circle
Given $\dfrac{dr}{dt} = 5$ Want $\dfrac{dA}{dt}$
$A = \pi r^2$
Find $\dfrac{dA}{dr} = 2\pi r$
$\dfrac{dA}{dt} = \dfrac{dr}{dt} \times \dfrac{dA}{dr} = 5(2\pi r) = 10\pi r$
$r = 8 \qquad \dfrac{dA}{dt} = 80\pi \;\text{m}^2/\text{s}$
A circle is increasing at rate of $6\,\text{m}^2/\text{s}$. Find rate of increase of radius when $r = 5\,\text{m}$.
Let $\dfrac{dA}{dt} = 6$ Need $\dfrac{dr}{dt}$
$A = \pi r^2$
$\dfrac{dA}{dr} = 2\pi r$
$\dfrac{dr}{dt} = \dfrac{dA}{dt} \div \dfrac{dA}{dr} = \dfrac{6}{2\pi r}$
$r = 8 \qquad \dfrac{3}{8\pi} \;\text{m/s}$
Units are important.
Section 3 of 6
Spheres
A sphere is increasing at rate of $12\,\text{m}^3/\text{s}$. Find rate of increase of radius when $r = 10\,\text{m}$.
$\dfrac{dV}{dt} = 12 \qquad \dfrac{dr}{dt} = ?$
$V = \dfrac{4}{3}\pi r^3$
$\dfrac{dV}{dr} = 4\pi r^2$
$\dfrac{dr}{dt} = \dfrac{dV}{dt} \div \dfrac{dV}{dr} = \dfrac{12}{4\pi r^2} = \dfrac{3}{\pi r^2}$
$r = 10 \qquad \text{Ans} \quad \dfrac{3}{100\pi} \;\text{m/s}$
A sphere is increasing at rate of $\dfrac{\pi}{1.9\times 10^{8}}\;\text{m}^2/\text{s}$. Find rate of increase of radius when $r = 2\,\text{m}$.
Sphere
$\dfrac{dA}{dt} = \dfrac{\pi}{1.9\times 10^{8}} \qquad$ Need $\dfrac{dr}{dt} = ?$
$A = 4\pi r^2$
$\dfrac{dA}{dr} = 8\pi r$
$\dfrac{dr}{dt} = \dfrac{dA}{dt} \div \dfrac{dA}{dr} = \dfrac{\frac{\pi}{1.9\times 10^{8}}}{8\pi r} = \dfrac{\pi}{1.9\times 10^{8}} \cdot \dfrac{1}{8\pi r}$
$r = 2 \qquad = \dfrac{1}{3.04\times 10^{9}} \;\text{m/s}$
Section 4 of 6
Cube and cylinder
A cube is melting a rate of $5\,\text{m}^3/\text{s}$. Find rate of decrease of sides.
$\dfrac{dV}{dt} = 5 \qquad$ Need $\dfrac{dx}{dt} = ??$
$V = x^3$
$\dfrac{dV}{dx} = 3x^2$
$\dfrac{dx}{dt} = \dfrac{dV}{dt} \div \dfrac{dV}{dx} = \dfrac{5}{3x^2} \;\text{m/s}$
A cylinder is increasing at rate of $4\,\text{m/s}$. Given the height is twice the radius find rate of increase of volume when $r = 3\,\text{m}$.
Let $\dfrac{dr}{dt} = 4 \qquad$ Need $\dfrac{dV}{dt}$
Find $V = \pi r^2 h \qquad h = 2r$
$V = 2\pi r^3$
$\dfrac{dV}{dr} = 6\pi r^2$
$\dfrac{dV}{dt} = \dfrac{dr}{dt} \times \dfrac{dV}{dr} = 4(6\pi r^2) = 24\pi r^2$
$r = 3 \qquad \dfrac{dV}{dt} = 216\pi \;\text{m}^3/\text{s}$
Section 5 of 6
Distance, speed, acceleration
Distance $= s = f(t) \qquad$ Units m
Distance is a function of time.
Speed $=$ change in distance with respect a change in time
$= \dfrac{ds}{dt} = f'(t) \qquad$ Units m/h or km/h or $\text{m/s} = \text{m}^{-1}$
Acceleration $= a =$ change in speed with a change in time.
$\dfrac{d}{dt}\dfrac{ds}{dt} = \dfrac{d^2 s}{dt^2} = f''(t) \qquad \text{m/s}^2 = \text{m}^{-2}$
Initial $\Rightarrow t = 0$
At rest $= \dfrac{ds}{dt} = 0$. Speed $= 0$
Distance $s$ travelled by a car in km over a time is seconds is given by $s = t^3 + 2t^2 + 5t + 1$. Find (i) Initial distance (ii) Speed after 2 sec (iii) Acceleration after 3 sec.
$s = t^3 + 2t^2 + 5t + 1$
(i) $t = 0 \qquad s = 1\;\text{km}$.
(ii) $\dfrac{ds}{dt} = 3t^2 + 4t + 5$
$t = 2 \qquad \dfrac{ds}{dt} = 3(2)^2 + 4(2) + 5 = 25\;\text{km/s}$
(iii) $\dfrac{d^2 s}{dt^2} = 6t + 4$
$t = 3 \qquad \dfrac{d^2 s}{dt^2} = 22\;\text{km/s}^2$
Section 6 of 6
Braking distance
The distance $s$ in m travelled by car in $t$ seconds after brakes are applied is given by $s = 12 + 8t - t^2$. Find distance travelled when at rest?
$\dfrac{ds}{dt} = 8 - 2t = 0$
$t = 4$
$s = 12 + 8(4) - 4^2 = 28\;\text{m}$
SUM
The lot in one box
Rate units
1.$\text{m/s} = \text{m}^{-1} = \dfrac{dr}{dt}$ = change in radius with respect to time.
2.$\text{m}^2/\text{s} = \text{m}^{-2} = \dfrac{dA}{dt}$
3.$\text{m}^3/\text{s} = \text{m}^{-3} = \dfrac{dV}{dt}$
End of lesson
Rates of change — HL · Mathslive.ie