Calculus: Derivatives IV

Related Rates

If the derivative of a function gives its rate of increase or decrease, then related rates are derivatives of two functions that are related in some way. For example, consider the rate at which a tankĀ is filled with water. The rate at which the water level rises is directly related to the rate at which water is entering or leaving the tank. Therefore, these two rates are said to be related.

Solving related rate problems requires understanding the relationships between two related functions and their related derivatives. In the above example, the two functions in question are the volume of water in the tank and the water level. Their derivatives are the flow of water into or out of the tank and the change in water level, respectively. Taking the derivatives is the easy part - figuring out how the two rates are related is the meaty part.


  1. One side of a right triangle is 10 units long. The other side starts at 5 units long and increases at a rate of 4 units per second. How fast is the area of the triangle increasing? (Neither side is the hypotenuse.)
    The area of a triangle is $A=\frac{1}{2}bh$. We let the height be the 10-unit side and the base be expressed as $5+4t$. Substituting these values into area formula gives us a function for the area of the triangle in terms of time:

    $A(t) = \dfrac{1}{2}10 (5+4t) \\ A(t) = 5(5+4t) \\ A(t) = 20t + 25$

    To get the rate at which the area is increasing, we simply take the derivative:

    $A'(t) = \dfrac{d}{dt}(20t+25) \\ A'(t) = 20$
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  2. One adjacent side of a right triangle is 8 units long. The other adjacent side starts at 4 units long and increases at a rate of 4 units per second. How fast is the length of the hypotenuse increasing?

    The Pythagorean Theorem relates the adjacent sides $a$ and $b$ of a right triangle to the hypotenuse $c$:

    $a^2 + b^2 = c^2$

    Solving for $c$ gives us:

    $c = \sqrt{a^2 + b^2}$

    Now we substitute in the values of each side to express the hypotenuse as a function of time.

    $c(t) = \sqrt{8^2 + (4+4t)^2} \\ c(t) = \sqrt{64 + 16t^2 + 32t + 16} \\ c(t) = \sqrt{16t^2 + 32t + 80}$

    To get the rate at which the hypotenuse is increasing, we simply take the derivative:

    $c'(t) = \dfrac{d}{dt}\sqrt{16t^2 + 32t + 80} \\ c'(t) = \dfrac{d}{dt}\left(16t^2 + 32t + 80\right)^{1/2} \\ c'(t) = \dfrac{1}{2}\left(16t^2 + 32t + 80\right)^{-1/2}\dfrac{d}{dt}\left(16t^2 + 32t + 80\right) \\ c'(t) = \dfrac{1}{2}\left(16t^2 + 32t + 80\right)^{-1/2}(32t + 32) \\ c'(t) = \dfrac{16t+16}{\sqrt{16t^2 + 32t + 80}} \\ c'(t) = \dfrac{16t+16}{4\sqrt{t^2 + 2t + 5}} \\ c'(t) = \dfrac{4t+4}{\sqrt{t^2 + 2t + 5}} \\ $
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  3. Two top fuel funny cars are lined up on the drag strip. Vince's car starts accelerating at 120m/s per second immediately when the light turns green. Nikki's car starts accelerating at 140m/s per second after 0.5 seconds. At what rate is the distance between them changing once Nikki has hit the gas?

    Hint: The formula for position $p$ in terms of time $t$ given acceleration $a$ and velocity $v$ is $p=\frac{1}{2}at^2 + vt + p_0$, where $p_0$ is the starting position.
    Using the formula for position, we can find the position of each ship as a function of time:
    • $p_V(t) = \dfrac{1}{2}120t^2 + 0t + 0=60t^2$
    • $p_N(t) = \dfrac{1}{2}140(t-0.5)^2 + 0t + 0 = 70(t-0.5)^2$
    We can can model the distance between Vince and Nikki's cars by taking the absolute value of the difference between their two displacements.

    $D(t) = \left| p_V(t) - p_N(t) \right| \\ D(t) = \left| 60t^2 - 70(t-0.5)^2 \right| \\ D(t) = \left| 60t^2 - 70(t^2-t+0.25) \right| \\ D(t) = \left| 60t^2 - 70t^2 + 70t - 17.5 \right| \\ D(t) = \left| -10t^2 + 70t - 17.5 \right| \\$

    Now we just differentiate to find the rate at which the distance is changing. This is also called the relative velocity.

    $D'(t) = \dfrac{d}{dt} \left| -10t^2 + 70t - 17.5 \right| \\ D'(t) = \left| -20t + 70 \right| \\$
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  4. Two ships leave port at the same time traveling in different directions. Ship A moves north at a constant rate of 5 knots, while Ship B accelerates east at a rate of 2 knots per hour. At what rate is the distance between the two ships increasing?

    Hint: The formula for position $p$ in terms of time $t$ given acceleration $a$ and velocity $v$ is $p=\frac{1}{2}at^2 + vt + p_0$, where $p_0$ is the starting position.
    Using the formula for position, we can find the position of each ship as a function of time:
    • $p_A(t) = \dfrac{1}{2}0t^2 + 5t + 0=5t$
    • $p_B(t) = \dfrac{1}{2}2t^2 + 0t + 0 = t^2$
    We can model the distance between the two ships as the hypotenuse of a right triangle. From the Pythagorean Theorem, we start with:

    $C = \sqrt{A^2 + B^2}$

    Now we can substitute our values back in:

    $C(t) = \sqrt{(5t)^2 + \left(t^2\right)^2} \\ C(t) = \sqrt{25t^2 + t^4} \\ $

    Now we differentiate to find the rate at which they're moving apart:

    $C'(t) = \dfrac{d}{dt}\sqrt{25t^2 + t^4} \\ C'(t) = \dfrac{d}{dt}\left(25t^2 + t^4\right)^{1/2} \\ C'(t) = \dfrac{1}{2}\left(25t^2 + t^4\right)^{-1/2}\dfrac{d}{dt}\left(25t^2 + t^4\right) \\ C'(t) = \dfrac{1}{2}\left(25t^2 + t^4\right)^{-1/2}\left(50t + 4t^3\right) \\ C'(t) = \dfrac{25t + 2t^3}{\sqrt{25t^2 + t^4}} \\ $

    Since $t$ is always positive in this scenario, we can simplify a bit more:

    $C'(t) = \dfrac{25t + 2t^3}{|t|\sqrt{25 + t^2}} \\ C'(t) = \dfrac{25 + 2t^2}{\sqrt{25 + t^2}} \\ $
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  5. A 10,000 cubic foot pool is being filled at a rate of 20 cubic feet per minute. The pool is 10 feet deep everywhere. At what rate is the water level rising?

    Let $V(t)$ be the amount of water in the pool at time $t$, and let $h(t)$ be the height of the water in the pool at time $t$. Given $V'(t)=20$, we need to solve for $h'(t)$.

    We first need to relate the volume of the pool and its height. Volume equals height times surface area:

    $V = hc$, where $V$ is the volume of the pool, $h$ is the height, and $c$ is some constant representing the surface area of the pool.

    Solving for $c$ will give us the square footage of the pool:

    $10,000 = 10c \\ c = 1,000$

    Now we can solve for the rate at which the water level is rising:

    $V(t) = 1,000h(t) \\ V'(t) = 1,000h'(t) \\ 20 = 1,000h'(t) \\ h'(t) = .02 \\ $

    The water level of the pool is rising $0.02$ feet per minute.
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  6. A 40' ladder is leaning against a wall. The bottom of the ladder starts moving away from the wall at a rate of 2'/s. How fast is the top of the ladder moving when the bottom of the ladder is 20' away from the wall?

    We can model the situation as a right triangle, where the hypotenus is 40', and the two adjacent sides are functions of time. Model these sides as $A(t)$ for the height and $B(t)$ for the base:

    $A^2(t) + B^2(t) = 40^2 \\ A^2(t) + B^2(t) = 1600 \\ $

    We're trying to find $A'(t)$, so we should differentiate both sides with respect to $t$:

    $\dfrac{d}{dt}A^2(t) + \dfrac{d}{dt}B^2(t) = \dfrac{d}{dt}1600 \\ 2A(t)\dfrac{dA}{dt}(t) + 2B(t)\dfrac{dB}{dt}(t) = 0 \\ A(t)\dfrac{dA}{dt}(t) + B(t)\dfrac{dB}{dt}(t) = 0 \\ $

    From the problem statement, we know that $\dfrac{dB}{dt}=2$:

    $A(t)\dfrac{dA}{dt}(t) + 2B(t) = 0 \\$

    When $B(t)=20$, we can solve for $A(t)$ with the Pythagorean theorem. Notice that we don't need to solve for $t$ explicitly. However, we should denote this specific time as $t_0$ instead of using the variable $t$.

    $A^2(t_0) + 20^2 = 1600 \\ A^2(t_0) + 400 = 1600 \\ A^2(t_0) = 1200 \\ A(t_0) = \sqrt{1200} \\ A(t_0) = 20\sqrt{3} \\ $

    Substituting everything back into the previous equation gives us:

    $20\sqrt{3}\dfrac{dA}{dt}(t_0) + 2\cdot20 = 0 \\ 20\sqrt{3}\dfrac{dA}{dt}(t_0) = - 40 \\ \dfrac{dA}{dt}(t_0) = -\dfrac{2}{\sqrt{3}} \\ $

    The top of the ladder is moving at $\frac{-2}{\sqrt{3}}$ feet per second.

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  7. Liquid chocolate is being pushed into a gigantic spherical balloon (apparently for good reason) at a rate of $10\text{m}^3/s$. How fast is the radius of the balloon expanding? Assume that the balloon starts off empty.

    Let the volume of the balloon be denoted by $V(t)=10t$.

    The change in volume with respect to time is therefore $V'(t) = 10$.

    The relationship between a sphere's radius and its volume is given by $V(t) = \dfrac{4}{3}\pi r^3(t)$. We should be able to differentiate both sides and solve for $r'(t)$:

    $V'(t) = \dfrac{4}{3} \pi 3r^2(t)r'(t) \\ V'(t) = 4\pi r^2(t)r'(t) \\ 10 = 4\pi r^2(t)r'(t) \\ r'(t) = \dfrac{5}{2\pi r^2(t)}$

    At this point it looks like we're in a pickle - what is $r(t)$? We can solve for it using the original equation for volume:

    $V(t) = \dfrac{4}{3}\pi r^3(t) \\ \dfrac{3}{4\pi}V(t) = r^3(t) \\ r(t) = \left({\dfrac{3}{4\pi}V(t)}\right)^{1/3} \\ r(t) = \left(\dfrac{3}{4\pi}10t\right)^{1/3} \\ r(t) = \left(\dfrac{15t}{2\pi}\right)^{1/3}$

    Plugging back into the equation for $r'(t)$ we get:

    $r'(t) = \dfrac{5}{2\pi \left(\left(\dfrac{15t}{2\pi}\right)^{1/3}\right)^2} \\ r'(t) = \dfrac{5}{2\pi \left(\dfrac{15t}{2\pi}\right)^{2/3}} \\ r'(t) = \dfrac{5}{2\pi} \left(\dfrac{2\pi}{15t}\right)^{2/3} \\ $
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  8. A giant ice cream cone is being filled with soft-serve ice cream at a rate of two cubic feet per minute. The ice cream cone is 4 feet deep and 8 feet in diameter. If the cone starts off empty, how fast is the height of the ice cream rising?

    Let the volume of the ice cream cone be denoted by $V(t) = 2t$.

    The change in volume with respect to time is accordingly $V'(t) = 2$.

    The formula for the volume of a cone in terms of its radius and height is given by $V = \frac{1}{3}\pi r^2 h$. Here, we know that the height, radius, and volume will all depend on time, so we adjust the formula to reflect this:

    $V(t) = \frac{1}{3} \pi r^2(t) h(t)$

    Don't differentiate just yet! You can differentiate now and still solve for the right answer, but it will be a big pain. First, we should simplify our equation. We have $r(t)$ and $h(t)$ expressed as two functions of time. However, we should be able to solve for one in terms of the other because we know the proportions of the cone.

    We will use similar triangles. Consider a cross section of the cone. Notice that $r$ and $h$ are both adjacent sides of a right triangle whose hypotenuse is the wall of the cone. We know that when height is 4, the radius is 4. Because side lengths have the same proportions in similar triangles, we have

    $\dfrac{r(t)}{h(t)} = \dfrac{4}{4} \\ r(t) = h(t)$

    Well, that was easy. Let's substitute this back into our earlier equation to simplify things:

    $V(t) = \dfrac{1}{3}\pi h^3(t)$

    Now we can differentiate:

    $V'(t) = \pi h^2(t)h'(t)\\$

    Now we plug in the value for $V'(t)$ and solve for $h'(t)$:

    $2 = \pi h^2(t)h'(t) \\ h'(t) = \dfrac{2}{\pi h^2(t)}$

    Now we just need to solve for $h(t)$. We can do this using the equation for $V(t)$:

    $V(t) = \dfrac{\pi}{3}h^3(t) \\ 2t = \dfrac{\pi}{3}h^3(t) \\ \dfrac{6t}{\pi} = h^3(t) \\ \left(\dfrac{6t}{\pi}\right)^{2/3} = h^2(t) \\$

    Our final equation is now:

    $h'(t) = \dfrac{2}{\pi \left(\frac{6t}{\pi}\right)^{2/3}} \\ h'(t) = \dfrac{2}{\pi}\left(\dfrac{\pi}{6t}\right)^{2/3}$
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  9. A test missile flies horizontally across the sky at 500mph at an altitude of 1 mile. A laser gun on the ground fires a continuous laser at the missile to try to shoot it down. If the missile is directly above the laser at time $t=0$, how fast is the angle between the laser and the ground changing?

    We can model this scenario using a right triangle. Let one adjacent side be the altitude of the missile, and the other adjacent side be its horizontal distance from the laser gun, and let the laser beam be the hypotenuse, which we'll label $c$. We want to solve for the angle between the laser beam and the ground, which we will label $\phi$. Let's also label the angle between the laser gun and the sky as $\theta$. Here's a picture:

    Diagram showing the relationship between the laser and the missile

    First, let's solve for $c$ using the Pythagorean theorem:

    $c^2 = 1^2 + (500t)^2 \\ c^2 = 1 + 250000t^2 \\ c = \sqrt{1 + 250000t^2}$

    By the law of sines,

    $\dfrac{500t}{\sin(\theta)} = \dfrac{\sqrt{1+250000t^2}}{\sin\left(\frac{\pi}{2}\right)} \\ \dfrac{500t}{\sin(\theta)} = \sqrt{1+250000t^2} \\ \dfrac{500t}{\sqrt{1+250000t^2}} = \sin(\theta) \\ \theta = \sin^{-1}\left(\dfrac{500t}{\sqrt{1+250000t^2}}\right) \\ $

    To get the angle between the ground and the laser, $\phi$, we simply subtract $\theta$ from $\frac{\pi}{2}$:

    $\phi = \dfrac{\pi}{2} - \sin^{-1}\left(\dfrac{500t}{\sqrt{1+250000t^2}}\right)$

    Finally, we differentiate to find the rate at which the angle is changing:

    $\phi' = -\dfrac{1}{\sqrt{1-\left(\dfrac{500t}{\sqrt{1+250000t^2}}\right)^2}}\cdot \dfrac{d}{dt}\dfrac{500t}{\sqrt{1+250000t^2}} \\ \phi' = -\dfrac{1}{\sqrt{1-\dfrac{250000t^2}{1+250000t^2}}}\cdot \dfrac{\sqrt{1+250000t^2}500 - 500t\frac{250000t}{\sqrt{1+250000t^2}}}{1+250000t^2} \\ \phi' = -\dfrac{1}{\sqrt{1-\dfrac{250000t^2}{1+250000t^2}}}\cdot 500\dfrac{\sqrt{1+250000t^2} - \frac{250000t^2}{\sqrt{1+250000t^2}}}{1+250000t^2} \\ \phi' = -\dfrac{1}{\sqrt{1-\dfrac{250000t^2}{1+250000t^2}}}\cdot 500\dfrac{\frac{1+250000t^2}{\sqrt{1+250000t^2}} - \frac{250000t^2}{\sqrt{1+250000t^2}}}{1+250000t^2} \\ \phi' = -\dfrac{1}{\sqrt{1-\dfrac{250000t^2}{1+250000t^2}}}\cdot 500\dfrac{\frac{1}{\sqrt{1+250000t^2}}}{1+250000t^2} \\ \phi' = -\dfrac{1}{\sqrt{\dfrac{1+250000t^2}{1+250000t^2}-\dfrac{250000t^2}{1+250000t^2}}}\cdot 500\dfrac{\frac{1}{\sqrt{1+250000t^2}}}{1+250000t^2} \\ \phi' = -\dfrac{1}{\sqrt{\dfrac{1}{1+250000t^2}}}\cdot 500\dfrac{\frac{1}{\sqrt{1+250000t^2}}}{1+250000t^2} \\ \phi' = -\sqrt{1+250000t^2}\cdot 500\dfrac{\frac{1}{\sqrt{1+250000t^2}}}{1+250000t^2} \\ \phi' = -\dfrac{500}{1+250000t^2} \\$
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  10. A fairy is sprinkling fairy dust into a hemispherical wok of radius 64cm at a rate of 4 cubic centimeters per second. Assuming that the fairy uniformly sprinkles the fairy dust so that it does not build up in the center, how fast is the fairy dust level rising when the height is 32cm?

    Hint: The volume of a portion of a sphere of radius $r$ filled to a height of $h$ from the bottom is given by $V=\pi\left(rh^2 - \frac{1}{3}h^3\right)$.

    Let the volume of the wok be given by $V(t) = 4t$

    The change in volume with respect to time is thus $V'(t) = 4$.

    Modifying the equation for a partially filled sphere to depend on time, we get

    $V(t) = \pi\left(r(t)h^2(t) -\dfrac{1}{3}h^3(t)\right)$

    But don't differentiate yet! $r$ and $h$ are given as functions in terms of time, but if we know one then we should be able to figure out the other.

    We will use our knowledge of spheres and circles to solve for $h$ as a function of $r$. Consider a vertical cross-section of the hemisphere, and label the horizontal axis as $r$ and the vertical axis as $h$:

    Diagram of the wok

    The equation for a circle centered at $(a,b)$ of radius $p$ is given by $y=\pm\sqrt{p^2-(x-a)^2}+b$. Since our axes are $r$ and $h$, this gives us $h = \pm\sqrt{p^2-(r-a)^2} + b$. We know that the radius is 64, so we can fill that in first:

    $h = \pm\sqrt{64^2-(r-a)^2}+b \\ h = \pm\sqrt{4096-(r-a)^2}+b$

    Then, note that we're only interested in the bottom hemisphere, so we should use the negative half of the equation:

    $h = -\sqrt{4096 - (r-a)^2} + b$

    Then, we should center our circle ad $(0,64)$ so that when $r=0$, $h=0$ and when $r=64$, $h=64$.

    $h = -\sqrt{4096 - r^2} + 64$

    This gives us the height of the dust based on the radius. We want the radius in terms of height, so we should solve for $r$:

    $h = -\sqrt{4096 - r^2} + 64 \\ h - 64 = -\sqrt{4096 - r^2} \\ 64 - h = \sqrt{4096 - r^2} \\ (64-h)^2 = 4096 - r^2 \\ r^2 = 4096 - (64-h)^2 \\ r = \sqrt{4096 - (64-h)^2} \\$

    Now we can substitute this back into the earlier equation:

    $V(t) = \pi\left(\sqrt{4096 - (64-h(t))^2}h^2(t) -\dfrac{1}{3}h^3(t)\right)$

    Now we need to differentiate with respect to time and solve for $h'(t)$. Surprisingly, this will produce a clean solution. Just kidding.

    $V'(t) = \pi\left(\dfrac{d}{dt}\sqrt{4096 - (64-h(t))^2}h^2(t) - \dfrac{1}{3}\dfrac{d}{dt}h^3(t)\right) \\ 2 = \pi\left(\dfrac{\dfrac{d}{dt}(4096 - (64-h(t))^2)}{2\sqrt{4096 - (64-h(t))^2}}h^2(t) + \sqrt{4096 - (64-h(t))^2}2h(t)h'(t) - h^2(t)h'(t)\right) \\ 2 = \pi\left(\dfrac{ - 2(64-h(t))(-h'(t))}{2\sqrt{4096 - (64-h(t))^2}}h^2(t) + \sqrt{4096 - (64-h(t))^2}2h(t)h'(t) - h^2(t)h'(t)\right) \\ 2 = \pi\left(\dfrac{ (64-h(t))h'(t)}{\sqrt{4096 - (64-h(t))^2}}h^2(t) + \sqrt{4096 - (64-h(t))^2}2h(t)h'(t) - h^2(t)h'(t)\right)$

    At some time $t_0$, the fairy dust level $h(t_0)$ will be 32cm. We need to find $h'(t_0)$, the rate at which the fairy dust level is rising at that moment.

    $2 = \pi\left(\dfrac{ (64-32)h'(t_0)}{\sqrt{4096 - (64-32)^2}}32^2 + \sqrt{4096 - (64-32)^2}2(32)h'(t_0) - 32^2h'(t_0)\right) \\ 2 = \pi\left(\dfrac{ 32h'(t_0)}{\sqrt{4096 - 1024}}1024 + \sqrt{4096 - 1024}64h'(t_0) - 1024h'(t_0)\right) \\ 2 = \pi\left(\dfrac{ 32h'(t_0)}{32\sqrt{3}}1024 + 2048\sqrt{3}h'(t_0) - 1024h'(t_0)\right) \\ 2 = \pi\left(\dfrac{ h'(t_0)}{\sqrt{3}}1024 + 2048\sqrt{3}h'(t_0) - 1024h'(t_0)\right) \\ 2 = \pi h'(t_0)\left(\dfrac{ 1024}{\sqrt{3}} + 2048\sqrt{3} - 1024\right) \\ 2 = \pi h'(t_0)\left(\dfrac{ 1024}{\sqrt{3}} + \dfrac{6114}{\sqrt{3}} - \dfrac{1024\sqrt{3}}{\sqrt{3}}\right) \\ 2 = 1024\pi h'(t_0)\left(\dfrac{ 7-\sqrt{3}}{\sqrt{3}}\right) \\ 2 = 1024\pi h'(t_0)\left(\dfrac{ 7\sqrt{3}-3}{3}\right) \\ h'(t_0) = \dfrac{3}{512(7\sqrt{3}-3)\pi} \\$

    The fairy dust level is rising at $\frac{3}{512(7\sqrt{3}-3)\pi}$ cm per second.

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