Related rates: geometric models Problem set A One question at a time

The sides of a square sheet of metal are increasing at 3 cm s\(^{-1}\). Find the rate of increase of the area at the instant when the area is 64 cm².

• \(A=x^2\)
• \(\frac{dA}{dt}=2x\frac{dx}{dt}\)
• \(A=64\Rightarrow x=8\)
• \(\frac{dA}{dt}=2(8)(3)=48\)

An oil slick is a circle. When its radius is 5 m, the radius is increasing at 0.4 m per minute. Find the rate of increase of the area at this instant.

• \(A=\pi r^2\)
• \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\)
• \(=2\pi(5)(0.4)=12.6\)

A spherical balloon is inflated so that its volume increases at 60 cm³ s\(^{-1}\). Find the rate of increase of the radius at the instant when the volume is \(36\pi\) cm³.

• \(V=\frac43\pi r^3=36\pi\Rightarrow r=3\)
• \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\)
• \(60=4\pi(3)^2\frac{dr}{dt}\)
• \(\frac{dr}{dt}=0.531\)

A rectangular photo frame is changing size. Its width x cm increases at 3 cm s\(^{-1}\) and its height y cm decreases at 1 cm s\(^{-1}\). When \(x=6\) and \(y=8\), find

1. (a) the rate of increase of the area;

   • \(A=xy\)
   • \(\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\)
   • \(=(6)(-1)+(8)(3)=18\)

2. (b) the rate of increase of the diagonal.

   • \(D^2=x^2+y^2\)
   • \(D=10\)
   • \(\frac{dD}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{D}\)
   • \(=\frac{(6)(3)+(8)(-1)}{10}=1\)

A circular patch of damp on a wall has radius r cm and area A cm². At one moment the radius is increasing at 0.5 cm s\(^{-1}\) and the area is increasing at 20 cm² s\(^{-1}\). Find the radius at this moment.

• \(A=\pi r^2\)
• \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\)
• \(20=2\pi r(0.5)\)
• \(r=\frac{20}{1\pi}=6.37\)

A 13 m ladder rests against a vertical wall on horizontal ground. The foot of the ladder slides away from the wall. When the foot is 5 m from the wall it is moving at 0.6 m s\(^{-1}\). Find the rate at which the top of the ladder is moving down the wall at that instant.

• \(x^2+y^2=169\Rightarrow y=(169-x^2)^{1/2}\)
• \(\frac{dy}{dt}=-\frac{x}{\sqrt{169-x^2}}\frac{dx}{dt}\)
• \(y=12:\ \frac{dy}{dt}=-\frac{5}{12}\times0.6=-0.25\)
• The negative sign means the top is moving down.

A kite flies at a constant height of 30 m above a student's eye level. It moves horizontally at 4 m s\(^{-1}\), directly away from the student. Find the rate at which the distance D between the student and the kite is increasing when the kite is 40 m horizontally from the student.

• \(D^2=x^2+900\)
• \(2D\frac{dD}{dt}=2x\frac{dx}{dt}\)
• \(D=50:\ \frac{dD}{dt}=\frac{x}{D}\frac{dx}{dt}=\frac{40}{50}\times4=3.2\)

A conical filter funnel (point down) has top diameter 8 cm and height 10 cm. Liquid runs in at 8 cm³ s\(^{-1}\).

1. (a) Show that \(V=\frac{4\pi}{75}h^3\), where V cm³ is the volume of water when its depth is h cm.

   • \(\frac{r}{h}=\frac{4}{10}\Rightarrow r=\frac{2h}{5}\)
   • \(V=\frac13\pi r^2h=\frac13\pi\left(\frac{2h}{5}\right)^2h\)
   • \(V=\frac{4\pi}{75}h^3\)

2. (b) Find the rate at which the depth is increasing when \(h=5\).

   • \(\frac{dV}{dh}=\frac{4\pi}{25}h^2\)
   • \(\frac{dV}{dt}=\frac{4\pi}{25}h^2\frac{dh}{dt}\)
   • \(8=4\pi\frac{dh}{dt}\)
   • \(\frac{dh}{dt}=0.637\)

A tracker on the ground is 500 m from the base of a weather balloon launch. The balloon rises vertically. When the balloon is 650 m up it is rising at 60 m s\(^{-1}\). Find the rate at which the angle of elevation \(\theta\) is increasing at that moment. Give your answer in radians per second to 3 significant figures.

• \(\tan\theta=\frac{h}{500}\), so \(\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{500}\frac{dh}{dt}\)
• \(\sec^{2}\theta=1+\left(\frac{650}{500}\right)^{2}=2.69\)
• \(\frac{d\theta}{dt}=\frac{60}{500\times2.69}=0.0446\)

The density of a gas leaking from a sealed syringe is \(\rho=\frac{m}{V}\), where m g is the mass and V cm³ is the volume. When \(\rho=4\) g cm\(^{-3}\), the mass is decreasing at 3 g s\(^{-1}\) and the volume is decreasing at 1 cm³ s\(^{-1}\). Determine, with justification, whether the density is increasing or decreasing.

• \(\rho=\frac{m}{V}\)
• \(\frac{d\rho}{dt}=\frac{V\frac{dm}{dt}-m\frac{dV}{dt}}{V^2}=\frac{\frac{dm}{dt}-\rho\frac{dV}{dt}}{V}\)
• \(\frac{dm}{dt}-\rho\frac{dV}{dt}=(-3)-4(-1)=1\)
• V is positive and the top is positive, so the density is increasing.

A rigid 5 m metal rod slides with one end on the floor and the other against a wall. When the top end is 4 m above the floor it is moving down at 0.2 m s\(^{-1}\). Find the speed at which the lower end is moving away from the wall.

• \(x^2+y^2=25\Rightarrow x=(25-y^2)^{1/2}\)
• \(\frac{dx}{dt}=-\frac{y}{\sqrt{25-y^2}}\frac{dy}{dt}\)
• \(x=3:\ \frac{dx}{dt}=-\frac{4}{3}\times(-0.2)=0.267\)