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Suppose we are given one quantity $$$x$$$ that depends on another quantity $$$y$$$. If rate of change of quantity $$$x$$$ with respect to time $$$t$$$ is given, how do we find rate of change of $$$y$$$ with respect to time? We need to do it because in real-worlds problems it is often easier to calculate rate of change of $$$x$$$ then rate of change of $$$y$$$.

In fact it can be easily done using the chain rule and other differentiating rules.

Example 1. Air is being pumped into a spherical balloon so that its volume increases at a rate of 200 $$$\frac{{{c}{{m}}^{{3}}}}{{s}}$$$. How fast is the radius of the balloon increasing when the diameter is 60 cm?

Let $$${V}$$$ is the volume of balloon and $$${r}$$$ is its radius, then we need to find $$$\frac{{{d}{r}}}{{{d}{t}}}$$$ at $$${r}=\frac{{60}}{{2}}={30}$$$, given $$$\frac{{{d}{V}}}{{{d}{t}}}={200}$$$.

It is known that volume of spherical balloon is $$${V}=\frac{{4}}{{3}}\pi{{r}}^{{3}}$$$.

Differentiating with respect to $$${t}$$$ gives (using Chain Rule) $$$\frac{{{d}{V}}}{{{d}{t}}}=\frac{{{d}{V}}}{{{d}{r}}}\cdot\frac{{{d}{r}}}{{{d}{t}}}$$$.

Derivative of $$${V}$$$ with respect to $$${r}$$$ is $$$\frac{{{d}{V}}}{{{d}{r}}}={4}\pi{{r}}^{{2}}$$$.

So, $$$\frac{{{d}{V}}}{{{d}{t}}}={4}\pi{{r}}^{{2}}\frac{{{d}{r}}}{{{d}{t}}}$$$.

Plugging known quantities gives $$${200}={4}\pi\cdot{{30}}^{{2}}\frac{{{d}{r}}}{{{d}{t}}}$$$ or $$$\frac{{{d}{r}}}{{{d}{t}}}=\frac{{200}}{{{4}\pi\cdot{900}}}=\frac{{1}}{{{18}\pi}}$$$.

So, radius of balloon is increasing at the rate of $$$\frac{{1}}{{{18}\pi}}\ \frac{{{c}{m}}}{{s}}$$$.

Example 2. Car A is traveling west at 60 $$$\frac{{{m}{i}}}{{h}}$$$ and car B is traveling north at 40 $$$\frac{{{m}{i}}}{{h}}$$$. They are both approaching intersection of roads. At what rate are the cars approaching each other when car A is 0.4 mi and car B is 0.3 mi from the intersection?

Let C is intersection of roads.approaching cars rate of change

At a given time $$${t}$$$ let $$${x}$$$ be the distance from car A to C, let $$${y}$$$ be the distance from car B to C, and let $$${z}$$$ be the distance between the cars, where $$${x}$$$, $$${y},$$$ and $$${z}$$$ are measured in miles.

We are given that $$$\frac{{{d}{x}}}{{{d}{t}}}=-{60}$$$ and $$$\frac{{{d}{y}}}{{{d}{t}}}=-{40}$$$. (The derivatives are negative because $$${x}$$$ and $$${y}$$$ are decreasing) We are asked to find $$$\frac{{{d}{z}}}{{{d}{t}}}$$$.

$$${x}$$$, $$${y}$$$ and $$${z}$$$ are related by the Pythagorean Theorem:


Differentiating each side with respect to $$${t}$$$, we have following:

$$${2}{z}\frac{{{d}{z}}}{{{d}{t}}}={2}{x}\frac{{{d}{x}}}{{{d}{t}}}+{2}{y}\frac{{{d}{y}}}{{{d}{t}}}$$$ or $$$\frac{{{d}{z}}}{{{d}{t}}}=\frac{{1}}{{z}}{\left({x}\frac{{{d}{x}}}{{{d}{t}}}+{y}\frac{{{d}{y}}}{{{d}{t}}}\right)}$$$.

When $$${x}={0.4}$$$ and $$${y}={0.3}$$$ the Pythagorean Theorem tells that $$${z}=\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}=\sqrt{{{{\left({0.4}\right)}}^{{2}}+{{\left({0.3}\right)}}^{{2}}}}={0.5}$$$.

Therefore, $$$\frac{{{d}{z}}}{{{d}{t}}}=\frac{{1}}{{{0.5}}}{\left({0.4}\cdot{\left(-{60}\right)}+{0.3}\cdot{\left(-{40}\right)}\right)}=-{72}$$$.

The cars are approaching each other at a rate of $$$-{72}\frac{{{m}{i}}}{{h}}$$$.

Example 3. A water tank has the shape of an inverted circular cone with base radius 3 m and height 9 m. If water is being pumped into the tank at a rate of 2 $$$\frac{{{m}}^{{3}}}{\min}$$$, find the rate at which the water level is rising when the water is 4 m deep.circular tank rate

Let $$${V}$$$, $$${r}$$$, and $$${h}$$$ be the volume of the water, the radius of the surface, and the height at time $$${t}$$$, where $$${t}$$$ is measured in minutes.
We are given that $$$\frac{{{d}{V}}}{{{d}{t}}}={2}\frac{{{m}}^{{3}}}{\min}$$$ and we are asked to find $$$\frac{{{d}{h}}}{{{d}{t}}}$$$ when $$${h}$$$ is 4 m. The quantities $$${V}$$$ and $$${h}$$$ are related by the equation $$${V}=\frac{{1}}{{3}}\pi{{r}}^{{2}}{h}$$$.

From similarity of triangles we make conclusion that $$$\frac{{r}}{{3}}=\frac{{h}}{{9}}$$$ or $$${r}=\frac{{h}}{{3}}$$$.

Plugging into equation gives $$${V}=\frac{{1}}{{3}}\pi{{\left(\frac{{h}}{{3}}\right)}}^{{2}}{h}=\frac{{1}}{{27}}\pi{{h}}^{{3}}$$$.

Now differentiate both sides with respect to time $$${t}$$$:

$$$\frac{{{d}{V}}}{{{d}{t}}}=\frac{{{d}{V}}}{{{d}{h}}}\frac{{{d}{h}}}{{{d}{t}}}=\frac{{1}}{{9}}\pi{{h}}^{{2}}\frac{{{d}{h}}}{{{d}{t}}}$$$ or $$$\frac{{{d}{h}}}{{{d}{t}}}=\frac{{9}}{{\pi{{h}}^{{2}}}}\frac{{{d}{V}}}{{{d}{t}}}$$$.

Substituting given values gives $$$\frac{{{d}{h}}}{{{d}{t}}}=\frac{{9}}{{\pi\cdot{{4}}^{{2}}}}\cdot{2}=\frac{{9}}{{{8}\pi}}\approx{0.358}\frac{{m}}{\min}$$$.

Example 4. A man walks along a straight path at a speed of 3 $$$\frac{{{f{{t}}}}}{{s}}$$$. A searchlight is located on the ground 20 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight?

Let $$${x}$$$ be the distance from the man to the point on the path closest to the searchlight and $$${L}$$$ is the distance between searchlight and man. Let $$$\theta$$$ be the angle between the beam of the searchlight and the perpendicular to the path.

change of rate of searchlight

We are given that $$$\frac{{{d}{x}}}{{{d}{t}}}={3}$$$ and we need to find $$$\frac{{{d}\theta}}{{{d}{t}}}$$$ when $$${x}={15}$$$.

We need to connect $$${x}$$$ and $$$\theta$$$. This relation can be easily found. From right-angled triangle $$${\tan{{\left(\theta\right)}}}=\frac{{x}}{{20}}$$$.

Differentiating both sides with respect to time gives

$$${{\sec}}^{{2}}{\left(\theta\right)}\frac{{{d}\theta}}{{{d}{t}}}=\frac{{1}}{{20}}\frac{{{d}{x}}}{{{d}{t}}}$$$ or $$$\frac{{{d}\theta}}{{{d}{t}}}=\frac{{1}}{{20}}{{\cos}}^{{2}}{\left(\theta\right)}\frac{{{d}{x}}}{{{d}{t}}}$$$.

When $$${x}={15}$$$ the Pythagorean theorem tells that $$${L}=\sqrt{{{{20}}^{{2}}+{{x}}^{{2}}}}=\sqrt{{{{20}}^{{2}}+{{15}}^{{2}}}}={25}$$$.

Therefore $$${\cos{{\left(\theta\right)}}}=\frac{{20}}{{L}}=\frac{{20}}{{25}}=\frac{{4}}{{5}}$$$.

Thus, $$$\frac{{{d}\theta}}{{{d}{t}}}=\frac{{1}}{{20}}{{\left(\frac{{4}}{{5}}\right)}}^{{2}}\cdot{3}=\frac{{12}}{{125}}={0.096}$$$.

The searchlight is rotating at a rate of $$${0.096}\frac{{{r}{a}{d}}}{{s}}$$$.

Example 5. A ladder whose length is 5 feet rests against vertical wall. If the bottom of the ladder slides away from the wall at a rate of $$${2}\ \frac{{f{{t}}}}{{s}}$$$, how fast is the top of the ladder sliding down the wall when the bottom of ladder is 4 feet from the wall.rate of change of sliding ladder

Let $$${x}$$$ feet be the distance from the bottom of the ladder to the wall and $$${y}$$$ feet the distance from the top of the ladder to the ground.

We have that $$$\frac{{{d}{x}}}{{{d}{t}}}={2}$$$. We need to find $$$\frac{{{d}{y}}}{{{d}{t}}}$$$ when $$${x}={4}$$$.

$$${x}$$$ and $$${y}$$$ are connected by Pythagorean Theorem as follows: $$${{x}}^{{2}}+{{y}}^{{2}}={{5}}^{{2}}={25}$$$.

Differentiating this equality with respect to $$${t}$$$ gives the following:

$$${2}{x}\frac{{{d}{x}}}{{{d}{t}}}+{2}{y}\frac{{{d}{y}}}{{{d}{t}}}={0}$$$ or $$$\frac{{{d}{y}}}{{{d}{t}}}=-\frac{{x}}{{y}}\frac{{{d}{x}}}{{{d}{t}}}$$$.

From Pythagorean theorem when $$${x}={4}$$$ we have that $$${{y}}^{{2}}={25}-{{x}}^{{2}}={25}-{16}={9}$$$ or $$${y}={3}$$$.

So $$$\frac{{{d}{y}}}{{{d}{t}}}=-\frac{{4}}{{3}}\cdot{2}=-\frac{{8}}{{3}}\ \frac{{f{{t}}}}{{s}}$$$.