AI HL · AHL 2
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Related rates: ladders, cones and circles
AHL 2.6
A 6.5 m ladder leans on a wall. Its foot slides away at a steady 0.3 m/s. Drag the foot to choose the moment, and see how fast the top is falling then.
The ladder never changes length, so x² + y² = 6.5². Differentiating with respect to t and dividing by 2 gives x(dx/dt) + y(dy/dt) = 0. The arrows show the speeds: orange for the foot, teal for the top.
Drag the foot of the ladder.
A 6.5 m ladder rests against a vertical wall. Its foot slides away from the wall. At the instant when the foot is 2.5 m from the wall, it is moving at 0.3 m s\(^{-1}\). Find \(\frac{dy}{dt}\), the rate of change of the height y of the top of the ladder.
Have a go first. Then open the working one line at a time.
\(\frac{dx}{dt}=0.3\) at \(x=2.5\). Want \(\frac{dy}{dt}\).
The foot's speed is dx/dt. The top's rate is dy/dt.
\(x^2+y^2=6.5^2\), so \(y=(42.25-x^2)^{1/2}\)
Pythagoras. The ladder's length never changes.
\(\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}=-\frac{x}{\sqrt{42.25-x^2}}\times\frac{dx}{dt}\)
Differentiate before you substitute.
\(y=\sqrt{42.25-2.5^2}=6\), so \(\frac{dy}{dt}=-\frac{2.5}{6}\times0.3=-0.125\)
The negative sign: the top is moving down.
The top is moving down the wall at 0.125 m s⁻¹.
Speed is the size of the rate.
Water runs at 2 m³ per minute into a tank shaped like an inverted cone (point down). The cone has top radius 3 m and height 9 m. Find the rate at which the water level is rising when the water is 6 m deep.
Have a go first. Then open the working one line at a time.
\(\frac{dV}{dt}=2\). Want \(\frac{dh}{dt}\) at \(h=6\).
dV/dt is the rate of filling. The level rising is dh/dt.
\(\frac{r}{h}=\frac{3}{9}\), so \(r=\frac{h}{3}\)
Similar triangles: the water's cross-section has the same shape as the cone.
\(V=\frac13\pi r^2h=\frac13\pi\left(\frac h3\right)^2h=\frac{\pi h^3}{27}\)
Only h and V are left.
\(\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}\)
Differentiate with respect to h, then link to time.
\(2=\frac{\pi(6)^2}{9}\frac{dh}{dt}=4\pi\frac{dh}{dt}\), so \(\frac{dh}{dt}=0.159\)
The level rises at 0.159 m per minute.
A spherical bubble is blown so that its volume increases at 12 cm³ s\(^{-1}\). Find the rate of increase of the radius when the radius is 2 cm.
Have a go first. Then open the working one line at a time.
\(\frac{dV}{dt}=12\). Want \(\frac{dr}{dt}\) at \(r=2\).
The volume rate is given. The radius rate is wanted.
\(V=\frac43\pi r^3\), so \(\frac{dV}{dr}=4\pi r^2\)
The formula for a sphere is in the formula booklet.
\(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\)
Rearranged: \(\frac{dr}{dt}=\frac{dV}{dt}\div4\pi r^2\).
\(12=4\pi(2)^2\frac{dr}{dt}\), so \(\frac{dr}{dt}=0.239\)
The radius grows at 0.239 cm per second. A larger sphere grows more slowly for the same dV/dt.
A rectangle has sides x cm and y cm. At one moment \(x=15\), \(y=8\), \(\frac{dx}{dt}=3\) cm s\(^{-1}\) and \(\frac{dy}{dt}=2\) cm s\(^{-1}\). Find the rate of increase of the diagonal at this moment.
Have a go first. Then open the working one line at a time.
\(D^2=x^2+y^2\), so \(D=17\) at this moment
Pythagoras for the diagonal.
\(2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\)
Each squared quantity brings a chain rule factor.
\(\frac{dD}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{D}\)
Divide by 2D.
\(\frac{dD}{dt}=\frac{15(3)+8(2)}{17}=\frac{61}{17}=3.59\)
The diagonal is increasing at 3.59 cm s⁻¹.
The density of a gas is \(\rho=\frac{m}{V}\), where m g is its mass and V cm³ its volume. When \(\rho=3\), the mass is increasing at 6 g s\(^{-1}\) and the volume is increasing at 4 cm³ s\(^{-1}\). Is the density increasing or decreasing?
Have a go first. Then open the working one line at a time.
\(\frac{d\rho}{dt}=\frac{V\frac{dm}{dt}-m\frac{dV}{dt}}{V^2}\)
Mass and volume both change.
\(\frac{d\rho}{dt}=\frac{\frac{dm}{dt}-\rho\frac{dV}{dt}}{V}\)
Divide top and bottom by V. This removes the unknown m.
\(\frac{d\rho}{dt}=\frac{6-3(4)}{V}=\frac{-6}{V}\)
V is positive, so the sign is the sign of the top.
The top is negative, so \(\frac{d\rho}{dt}<0\): the density is decreasing.
The volume grows faster than the mass allows the density to stay up.
A circular ripple has radius r cm and area A cm². At one moment the radius is increasing at 0.8 cm s\(^{-1}\) and the area is increasing at 12 cm² s\(^{-1}\). Find the radius at this moment.
Have a go first. Then open the working one line at a time.
\(A=\pi r^2\)
Area of a circle.
\(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\)
Both rates are given, so r is the unknown.
\(12=2\pi r(0.8)\)
Put in dA/dt = 12 and dr/dt = 0.8.
\(r=\frac{12}{1.6\pi}=2.39\)
The radius is 2.39 cm.
A camera on the ground is 600 m from the launch pad of a rocket that rises vertically. The camera tilts up to follow the rocket. When the rocket is 800 m up it is rising at 80 m s\(^{-1}\). Find the rate at which the angle of elevation \(\theta\) of the camera is increasing at that moment.
Have a go first. Then open the working one line at a time.
\(\frac{dh}{dt}=80\) at \(h=800\). Want \(\frac{d\theta}{dt}\).
The camera stays 600 m from the pad. Only h and θ change.
\(\tan\theta=\frac{h}{600}\)
Opposite over adjacent in the right-angled triangle.
\(\sec^{2}\theta\,\frac{d\theta}{dt}=\frac{1}{600}\frac{dh}{dt}\)
The derivative of tan θ is sec²θ, and the chain rule adds dθ/dt.
\(\sec^{2}\theta=1+\tan^{2}\theta=1+\left(\frac{800}{600}\right)^{2}=\frac{25}{9}\)
Use the instant's values only now.
\(\frac{d\theta}{dt}=\frac{9}{25}\times\frac{80}{600}=0.048\) rad s⁻¹
The angle is increasing, at 0.048 radians per second.