Related rates: geometric models Problem set B Worked solutions

The sides of a square patch of ice on a pond are increasing at 1.5 cm s\(^{-1}\). Find the rate of increase of the area at the instant when the area is 49 cm².

• \(A=x^2\)
• \(\frac{dA}{dt}=2x\frac{dx}{dt}\)
• \(A=49\Rightarrow x=7\)
• \(\frac{dA}{dt}=2(7)(1.5)=21\)

A ripple on a pond is a circle. When its radius is 8 cm, the radius is increasing at 2.5 cm s\(^{-1}\). Find the rate of increase of the area at this instant.

• \(A=\pi r^2\)
• \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\)
• \(=2\pi(8)(2.5)=126\)

A spherical soap bubble is blown so that its volume increases at 45 cm³ s\(^{-1}\). Find the rate of increase of the radius at the instant when the volume is \(288\pi\) cm³.

• \(V=\frac43\pi r^3=288\pi\Rightarrow r=6\)
• \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\)
• \(45=4\pi(6)^2\frac{dr}{dt}\)
• \(\frac{dr}{dt}=0.0995\)

A rectangular screen is being resized. Its width x cm decreases at 2 cm s\(^{-1}\) and its height y cm increases at 4 cm s\(^{-1}\). When \(x=5\) and \(y=12\), find

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

   • \(A=xy\)
   • \(\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\)
   • \(=(5)(4)+(12)(-2)=-4\)

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

   • \(D^2=x^2+y^2\)
   • \(D=13\)
   • \(\frac{dD}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{D}\)
   • \(=\frac{(5)(-2)+(12)(4)}{13}=2.92\)

A circular stain on a carpet 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 30 cm² s\(^{-1}\). Find the radius at this moment.

• \(A=\pi r^2\)
• \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\)
• \(30=2\pi r(0.8)\)
• \(r=\frac{30}{1.6\pi}=5.97\)

A 10 m plank leans against a vertical fence on horizontal ground. The foot of the plank slides away from the fence. When the foot is 6 m from the fence it is moving at 0.5 m s\(^{-1}\). Find the rate at which the top of the plank is moving down the fence at that instant.

• \(x^2+y^2=100\Rightarrow y=(100-x^2)^{1/2}\)
• \(\frac{dy}{dt}=-\frac{x}{\sqrt{100-x^2}}\frac{dx}{dt}\)
• \(y=8:\ \frac{dy}{dt}=-\frac{6}{8}\times0.5=-0.375\)
• The negative sign means the top is moving down.

A drone flies at a constant height of 24 m above a camera on the ground. It moves horizontally at 5 m s\(^{-1}\), directly away from the camera. Find the rate at which the distance D between the camera and the drone is increasing when the drone is 7 m horizontally from the camera.

• \(D^2=x^2+576\)
• \(2D\frac{dD}{dt}=2x\frac{dx}{dt}\)
• \(D=25:\ \frac{dD}{dt}=\frac{x}{D}\frac{dx}{dt}=\frac{7}{25}\times5=1.4\)

A conical paper cup (point down) has top diameter 6 cm and height 12 cm. Water runs in at 9 cm³ s\(^{-1}\).

1. (a) Show that \(V=\frac{\pi}{48}h^3\), where V cm³ is the volume of water when its depth is h cm.

   • \(\frac{r}{h}=\frac{3}{12}\Rightarrow r=\frac{h}{4}\)
   • \(V=\frac13\pi r^2h=\frac13\pi\left(\frac{h}{4}\right)^2h\)
   • \(V=\frac{\pi}{48}h^3\)

2. (b) Find the rate at which the depth is increasing when \(h=6\).

   • \(\frac{dV}{dh}=\frac{\pi}{16}h^2\)
   • \(\frac{dV}{dt}=\frac{\pi}{16}h^2\frac{dh}{dt}\)
   • \(9=\frac{9\pi}{4}\frac{dh}{dt}\)
   • \(\frac{dh}{dt}=1.27\)

An observer stands 120 m from the foot of a tall crane. The crane's hook rises vertically. When the hook is 90 m up it is rising at 4 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.

• \(\tan\theta=\frac{h}{120}\), so \(\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{120}\frac{dh}{dt}\)
• \(\sec^{2}\theta=1+\left(\frac{90}{120}\right)^{2}=1.562\)
• \(\frac{d\theta}{dt}=\frac{4}{120\times1.562}=0.0213\)

The density of a sample of salt solution evaporating in a dish is \(\rho=\frac{m}{V}\), where m g is the mass and V cm³ is the volume. When \(\rho=6\) g cm\(^{-3}\), the mass is decreasing at 7 g s\(^{-1}\) and the volume is decreasing at 1 cm³ s\(^{-1}\). Determine, with justification, whether the density is increasing or decreasing.

• \(\rho=\frac{m}{V}\)
• \(\frac{d\rho}{dt}=\frac{V\frac{dm}{dt}-m\frac{dV}{dt}}{V^2}=\frac{\frac{dm}{dt}-\rho\frac{dV}{dt}}{V}\)
• \(\frac{dm}{dt}-\rho\frac{dV}{dt}=(-7)-6(-1)=-1\)
• V is positive and the top is negative, so the density is decreasing.

A 13 m pole is leaned against a wall and slides, with its foot on horizontal ground. When the top of the pole is 12 m above the ground it is moving down at 0.15 m s\(^{-1}\). Find the speed at which the foot of the pole is moving away from the wall.

• \(x^2+y^2=169\Rightarrow x=(169-y^2)^{1/2}\)
• \(\frac{dx}{dt}=-\frac{y}{\sqrt{169-y^2}}\frac{dy}{dt}\)
• \(x=5:\ \frac{dx}{dt}=-\frac{12}{5}\times(-0.15)=0.360\)