Related Rates

A sunken oil tanker is leaking oil in an expanding slick the radius of whose circle is increasing at the rate of $3\text{ m/s}$. What is the rate at which the area of the circle expands when the radius of the oil slick is $50\text{m}$?

We are filming a rocket launch with a camera installed $900$ m away from the launching pad. If the rocket is ascending at $400$ m/s when it is $1200$ m above the launching pad, how fast is the distance between the rocket and the camera changing?

Solution

Let $t=$ time in seconds since the launch $h=$ height of the rocket in meters after $t$ seconds $s=$ distance between the rocket and the camera in meters after $t$ seconds We want to find $${\frac{ds}{dt}|}_{h=1200}\text{given}{\frac{dh}{dt}|}_{h=1200}=400\text{ m/s}.$$

By the Pythagorean Theorem: (900)^{2}+h^{2}=s^{2}.\tag{i} Since $h$ and $s$ are functions of $t$, differentiating both sides with respect to $t$: $$0+2h\frac{dh}{dt}=2s\frac{ds}{dt}$$ \frac{ds}{dt}=\frac{h}{s}\frac{dh}{dt}.\tag{ii} When $h=1200$ m, it follows from (i) that $s=\sqrt{{900}^2+{1200}^2}=1500$ m. Substituting into (ii): $$\frac{ds}{dt}=\frac{1200}{1500}\cdot 400=320\frac{\text{m}}{\text{s}}.$$ The distance between the rocket and the camera is increasing at a rate of $320$ m/s.

Using the same setup as the previous example (camera $900$ m from the launch pad, rocket ascending at $400$ m/s when $h=1200$ m), what should the instantaneous rate of change of the camera's elevation angle be to keep the rocket in view?

Solution

Let $t=$ time in seconds since the launch $h=$ height of the rocket in meters after $t$ seconds $\theta =$ camera's elevation angle in radians after $t$ seconds We want to find $${\frac{d\theta }{dt}|}_{h=1200}\text{given}{\frac{dh}{dt}|}_{h=1200}=400\text{ m/s}.$$

From the figure: \tan\theta=\frac{h}{900}.\tag{i} Since $\theta$ and $h$ are functions of $t$, differentiating with respect to $t$: (\sec^{2}\theta)\frac{d\theta}{dt}=\frac{1}{900}\frac{dh}{dt}\tag{ii} \frac{d\theta}{dt}=\frac{1}{900\sec^{2}\theta}\frac{dh}{dt}.\tag{iii} When $h=1200$, by the Pythagorean Theorem $s=1500$ m and $$\sec \theta =\frac{1}{\cos \theta }=\frac{1500}{900}=\frac{5}{3}\text{when}h=1200\text{ m.}$$ Substituting into (iii): $${\frac{d\theta }{dt}|}_{h=1200}=\frac{1}{900{\left(\frac{5}{3}\right)}^2}\cdot 400=\frac{400}{900\cdot \frac{25}{9}}=\frac{400}{2500}=\frac{4}{25}=0.16\text{ rad/s.}$$ When $h=1200$ m and the speed of the rocket is $400$ m/s, the elevation angle must be increased at a rate of $0.16$ rad/s to keep the rocket in view.

A ladder $150$ cm long leaning against a wall slips, and its foot moves away from the wall at $60$ cm/s. When the foot's base is $120$ cm from the wall, how fast does the top of the ladder approach the ground?

Solution

Let $t=$ time in seconds since the ladder slipped $x=$ distance of the ladder's base from the wall in centimeters after $t$ seconds $y=$ distance of the ladder's top from the ground in centimeters after $t$ seconds We want to find $${\frac{dy}{dt}|}_{x=120}\text{given}{\frac{dx}{dt}|}_{x=120}=60\text{ cm/s}.$$

By the Pythagorean Theorem: x^{2}+y^{2}=(150)^{2}.\tag{i} Differentiating with respect to $t$: $$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$ \frac{dy}{dt}=\frac{-x}{y}\frac{dx}{dt}.\tag{ii} When $x=120$ cm, equation (i) gives $y=\sqrt{{150}^2-{120}^2}=90$ cm. Substituting: $${\frac{dy}{dt}|}_{x=120}=-\frac{120}{90}(60)=-80\text{ cm/s.}$$ The negative sign indicates that $y$ is decreasing, meaning the top of the ladder is moving closer to the ground at $80$ cm/s.

Liquid is being filtered through a conical funnel. The height of the funnel is $15$ cm and the radius of its base is $5$ cm. The liquid is filtering out at the constant rate of $3{\text{cm}}^3/\text{s}$. What is the rate at which the height of the liquid in the funnel is decreasing when the height of the liquid is $4$ cm?

A person 5 feet tall is approaching a 15-foot lamp post at the constant rate of 4 ft/s.

(a) How fast is the length of the person's shadow decreasing?

(b) How fast is the tip of the person's shadow moving toward the lamp post?

Solution

**(a)** Let $x=$ distance of the person from the foot of the lamp post $s=$ length of the person's shadow We need to find ${\displaystyle \frac{ds}{dt}}$ given ${\displaystyle \frac{dx}{dt}}=-4$ ft/s (negative because $x$ is decreasing).

By similar triangles (the person and the lamp post both cast shadows from the same light source): $$\frac{s}{x+s}=\frac{5}{15}$$ Wait, more directly, using similar triangles \triangle ABB' \sim \triangle A'B'C (where the lamp is at $C$, the person's head is at $A$, and the tip of the shadow is at B'): $$\frac{s}{x}=\frac{5}{10}\Rightarrow s=\frac{1}{2}x.$$ Differentiating with respect to $t$: $$\frac{ds}{dt}=\frac{1}{2}\frac{dx}{dt}=\frac{1}{2}(-4)=-2\text{ ft/s.}$$ The length of the person's shadow is decreasing at $2$ ft/s. Note that $ds/dt$ is constant, independent of the person's position. **(b)** The position of the tip of the shadow is $x+s$ from the lamp post. The rate at which the tip is moving is: $$\frac{d(x+s)}{dt}=\frac{dx}{dt}+\frac{ds}{dt}=-4+(-2)=-6\text{ ft/s.}$$ The tip of the shadow is moving toward the lamp post at $6$ ft/s.

In the previous example, let $\theta$ be the angle between the ray from the lamp post to the tip of the shadow and the lamp post. What is the rate at which $\theta$ is changing when the person is $8$ ft away from the lamp post?

Solution

We need to find $${\frac{d\theta }{dt}|}_{x=8}\text{given}\frac{dx}{dt}=-4\text{ ft/s.}$$

From the figure (the lamp is 15 ft tall and the shadow tip is at horizontal distance $x+s=x+x/2=3x/2$ from the base, but using the geometry with the base distance $x+s$ from the lamp post): Using the lamp height of $15$ ft and horizontal distance $x+s$ for the total horizontal extent, but the more direct relation uses just $x$ (the distance of the person) compared to the height difference of $15-5=10$ ft between the lamp top and the person's head. The angle $\theta$ satisfies: \tan\theta=\frac{x}{10}.\tag{i} Differentiating with respect to $t$: $$(1+{\tan }^2\theta )\frac{d\theta }{dt}=\frac{1}{10}\frac{dx}{dt}$$ $$\frac{d\theta }{dt}=\frac{1}{10}{\cos }^2\theta \frac{dx}{dt}.$$ From the right triangle with legs $10$ (vertical) and $x$ (horizontal): $$\cos \theta =\frac{10}{\sqrt{{10}^2+x^2}}.$$ Therefore: \begin{aligned} \frac{d\theta}{dt} &= \frac{1}{10}\left(\frac{10}{\sqrt{10^{2}+x^{2}}}\right)^{2}\frac{dx}{dt} \\ &= \frac{10}{10^{2}+x^{2}}\frac{dx}{dt}. \end{aligned} At $x=8$: $${\frac{d\theta }{dt}|}_{x=8}=\frac{10}{100+64}\times (-4)=-\frac{40}{164}=-\frac{10}{41}\text{ rad/s.}$$ Converting to degrees per second: $$-\frac{10}{41}\frac{\text{rad}}{\text{s}}\times \frac{180°}{\pi \text{rad}}=-\frac{1800}{41\pi }°/\text{s}\approx -13.97°/\text{s.}$$ Note that $d\theta /dt$ depends on $x$, while $ds/dt$ (the rate at which the shadow length changes) is constant.

A police cruiser is approaching a right-angled intersection from the north and a speeding car is moving toward the intersection from the west. The police determine by radar that when the distance between them and the car is $1$ mile, the distance is decreasing at $70$ mph. If the cruiser is moving at $30$ mph, what is the speed of the car?

Solution

Let $y=$ position of the cruiser (measured from the intersection, moving southward is positive) $x=$ position of the car (measured from the intersection) $s=$ distance between the cruiser and the car at time $t$ We want to find ${\displaystyle \frac{dx}{dt}}$ when $y=0.6$ miles, $s=1$ mile, ${\displaystyle \frac{dy}{dt}}=-30$ mph, and ${\displaystyle \frac{ds}{dt}}=-70$ mph. Both $dy/dt$ and $ds/dt$ are negative because $y$ and $s$ are decreasing.

By the Pythagorean Theorem: s^{2}=x^{2}+y^{2}.\tag{i} Differentiating with respect to $t$: 2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.\tag{ii} Given $s=1$ mile and $y=0.6$ miles, from (i): $x^2=1-0.36=0.64$. Since the car is on the negative $x$-axis, $x=-0.8$ miles. Substituting into (ii): $$2(1)(-70)=2(-0.8)\frac{dx}{dt}+2(0.6)(-30)$$ $$-140=-1.6\frac{dx}{dt}-36$$ $$-1.6\frac{dx}{dt}=-104$$ $$\frac{dx}{dt}=65\text{ mph.}$$ The car's speed at the moment of measurement is $65$ mph. The positive sign of $dx/dt$ confirms that $x$ is increasing (the car is moving away from the intersection in the positive $x$ direction).