Related Rates involve finding the rate at which one quantity is changing by relating it to to other quantities with known rates. They apply concepts that we have previously covered, such as Implicit Differentiation and Chain Rule, in order to find the results.
Steps for Finding Related Rates
- Identify what's given and what you need to find (pay attention to units)
- Find an equation to differentiate
- Identify what variables change with time \(\cfrac{d?}{dt}\) and which are actually constant (\(0\))
- Substitute all the known values after you determine the derivative
- Use another equation if there are still too many unknowns
Example
Air leaks out of a ballon at a rate of \(3\; \left[\cfrac{\text{ft}^3}{\text{min}}\right]\). How fast is the Surface Area shrinking when the radius is \(10\;[\text{ft}]\)? Note: \(SA = 4\pi r^2\) and \(V = \cfrac{4}{3}\pi r^3\).
First, we can determine that as air leaks from the balloon that \(V\) (Volume), \(r\) (radius) and \(SA\) (Surface Area) all change with time. They are all functions of time and can be written as \(V(t)\), \(r(t)\), and \(SA(t)\) respectively:
Their Related Rates are expressed as such:
\(V'(t) = \cfrac{dV}{dt}\)
\(SA'(t) = \cfrac{dSA}{dt}\)
\(r'(t) = \cfrac{dr}{dt}\)
Next, we can identify the rate at which the balloon loses Volume:
\(\cfrac{dV}{dt} = -3\;\left[\cfrac{\text{ft}^3}{\text{min}}\right]\)
We can also determine what we need to find:
\(\cfrac{dSA}{dt} = ?\) at \(r = 10\)
We can create a table listing all of the known values and what we still need to determine:
Variable |
Quantity |
Units |
Rate |
Rate Units |
Volume (V) |
\(V = \cfrac{4}{3}\pi r^3\) |
\([\text{ft}^3]\) |
\(\cfrac{dV}{dt} = -3\) |
\(\left[\cfrac{\text{ft}^3}{\text{min}}\right]\) |
Surface Area (SA) |
\(SA = 4\pi r^2\) |
\([\text{ft}^2]\) |
\(\cfrac{dSA}{dt} = ?\) |
\(\left[\cfrac{\text{ft}^2}{\text{min}}\right]\) |
Radius (r) |
\(r = 10\) |
\([\text{ft}]\) |
\(\cfrac{dSA}{dt} = ?\) |
\(\left[\cfrac{\text{ft}}{\text{min}}\right]\) |
Then, we can set the Surface Area, \(\cfrac{dSA}{dt}\), equal to its equation. We can then determine its derivative using Implicit Differentiation:
\(\cfrac{dSA}{dt} = 4\pi r^2\)
\(\cfrac{dSA}{dt} = (2)4\pi r \cdot \cfrac{dr}{dt}\)
\(\cfrac{dSA}{dt} = 8\pi r \cdot \cfrac{dr}{dt}\)
Since we don't have all the values necessary to determine \(\cfrac{dSA}{dt}\), we can turn to Volume's Related Rate to solve \(\cfrac{dr}{dt}\).
We can differentiate the original Volume equation using Implicit Differentiation:
\(V = \cfrac{4}{3}\pi r^3\)
\(V' = (3) \left(\cfrac{4}{3}\pi r^2\right)\cdot \cfrac{dr}{dt}\)
\(V' = 4\pi r^2\cdot \cfrac{dr}{dt}\)
After, we can set Volume's Related Rate equal to this equation. We can then simplify this expression to solve \(\cfrac{dr}{dt}\):
\(\cfrac{dV}{dt} = 4\pi r^2\cdot \cfrac{dr}{dt}\)
\(-3 = 4\pi (10)^2\cdot \cfrac{dr}{dt}\)
\(-3 = 400\pi \cdot \cfrac{dr}{dt}\)
\(\cfrac{dr}{dt} = \cfrac{-3}{400\pi}\)
We can now substitute the known values into the equation and simplify to determine the rate of change for the Surface Area:
\(\cfrac{dSA}{dt} = 8\pi (10) \left(\cfrac{-3}{400\pi}\right)\)
\(\cfrac{dSA}{dt} = \cfrac{-240\cancel{\pi}}{400\cancel{\pi}}\)
\(\cfrac{dSA}{dt} = -\cfrac{3}{5}\)
Finally, we can complete our table by filling in the missing values:
Variable |
Quantity |
Units |
Rate |
Rate Units |
Volume (V) |
\(V = \cfrac{4}{3}\pi r^3\) |
\([\text{ft}^3]\) |
\(\cfrac{dV}{dt} = -3\) |
\(\left[\cfrac{\text{ft}^3}{\text{min}}\right]\) |
Surface Area (SA) |
\(SA = 4\pi r^2\) |
\([\text{ft}^2]\) |
\(\cfrac{dSA}{dt} = -\cfrac{3}{5}\) |
\(\left[\cfrac{\text{ft}^2}{\text{min}}\right]\) |
Radius (r) |
\(r = 10\) |
\([\text{ft}]\) |
\(\cfrac{dSA}{dt} = \cfrac{-3}{400 \pi}\) |
\(\left[\cfrac{\text{ft}}{\text{min}}\right]\) |
Therefore, we can determine that the Surface Area shrinks at a rate of \(-\cfrac{3}{5}\;\left[\cfrac{\text{ft}^2}{\text{min}}\right]\).
A small balloon is released at a point \(150\;[\text{ft}]\) away from an observer, who is on level ground. If the balloon goes straight up at a rate of \(8\; \left[\cfrac{\text{ft}}{\text{sec}}\right]\).
- How fast is the distance from the observer to the balloon increasing when the balloon is \(50\) feet high?
- How fast is the angle of elevation increasing?
Show Answer
i. From the question description, we can identify that \(a\) (adjacent) remains constant, while \(b\) (opposite), \(c\) (hypoteneuse), and \(\theta\) change in relation to time.
Their Related Rates are expressed as such:
\(b' = \cfrac{db}{dt}\)
\(c' = \cfrac{dc}{dt}\)
\(\theta' = \cfrac{d\theta}{dt}\)
Next, we can identify the rate at which the balloon rises up:
\(\cfrac{db}{dt} = 8\; \left[\cfrac{\text{ft}}{\text{sec}}\right]\)
We can also determine what we need to find:
\(\cfrac{dc}{dt} = ?\) at \(b = 50\)
We can create a table listing all of the known values and what we still need to determine:
Variable |
Quantity |
Units |
Rate |
Rate Units |
Adjacent (a) |
\(a = 150\) |
\([\text{ft}]\) |
N/A |
N/A |
Opposite (b) |
\(b = 50\) |
\([\text{ft}]\) |
\(\cfrac{db}{dt} = 8\) |
\(\left[\cfrac{\text{ft}}{\text{sec}}\right]\) |
Hypoteneuse (c) |
\(c = \sqrt{a^2 + b^2}\) |
\([\text{ft}]\) |
\(\cfrac{dc}{dt} = ?\) |
\(\left[\cfrac{\text{ft}}{\text{sec}}\right]\) |
Then, we can determine \(c\) by using the Pythagorean Theorem:
\(a^2 + b^2 = c^2\)
\((150)^2 + (50)^2 = c^2\)
\(\sqrt{22500+2500} = \sqrt{c^2}\)
\(c = \sqrt{25000}\)
After, we can differentiate the Pythagorean Theorem with the known values substituted using Implicit Differentiation:
\(a^2 + b^2 = c^2\)
\((150)^2 + b^2 = c^2\)
\(0 + 2b\cdot\cfrac{db}{dt} = 2c\cdot\cfrac{dc}{dt}\)
Finally, we can simplify the equation by substituting the known values for the corresponding variables to determine the rate of change for the distance between the observer and the balloon:
\(2(50)(8) = 2(\sqrt{25000})\cdot\cfrac{dc}{dt}\)
\(\cfrac{dc}{dt} = \cfrac{800}{2(\sqrt{25000})}\)
\(\cfrac{dc}{dt} = \cfrac{400}{\sqrt{25000}} = 2.53\)
Finally, we can complete our table by filling in the missing values:
Variable |
Quantity |
Units |
Rate |
Rate Units |
Adjacent (a) |
\(a = 150\) |
\([\text{ft}]\) |
N/A |
N/A |
Opposite (b) |
\(b = 50\) |
\([\text{ft}]\) |
\(\cfrac{db}{dt} = 8\) |
\(\left[\cfrac{\text{ft}}{\text{sec}}\right]\) |
Hypoteneuse (c) |
\(c = \sqrt{25000}\) |
\([\text{ft}]\) |
\(\cfrac{dc}{dt} = 2.53\) |
\(\left[\cfrac{\text{ft}}{\text{sec}}\right]\) |
Therefore, we can determine that the distance increases at a rate of roughly \(2.53\; \left[\cfrac{\text{ft}}{\text{sec}}\right]\).
ii. First, we can create a table listing all of the known values and what we still need to determine:
Variable |
Quantity |
Units |
Rate |
Rate Units |
Adjacent (a) |
\(a = 150\) |
\([\text{ft}]\) |
N/A |
N/A |
Opposite (b) |
\(b = 50\) |
\([\text{ft}]\) |
\(\cfrac{db}{dt} = 8\) |
\(\left[\cfrac{\text{ft}}{\text{sec}}\right]\) |
Angle (θ) |
\(θ = \tan^{-1}\cfrac{b}{a}\) |
\([\text{deg}]\) |
\(\cfrac{dθ}{dt} = ?\) |
\(\left[\cfrac{\text{deg}}{\text{sec}}\right]\) |
We can also draw a diagram to help visualize the problem:
Next, we can determine the value of \(\theta\) at \(b = 50\). We can do this by using the inverse trig function:
\(\tan\theta = \cfrac{b}{a}\)
\(\tan\theta = \cfrac{50}{150}\)
\(\theta = \tan^{-1}\left(\cfrac{1}{3}\right)\)
\(\theta = 18^{\circ}\)
Then, we can differentiate the \(\tan\) equation using Implicit Differentiation with \(150\) substituted for \(a\):
\(\tan\theta = \cfrac{b}{150}\)
\(\sec^2\theta\cdot\cfrac{d\theta}{dt} = \cfrac{1}{150}\cdot\cfrac{db}{dt}\)
Finally, we can simplify by substituting the known values in order to determine the speed the angle of elevation is increasing:
\(\sec^218^{\circ}\left(\cfrac{d\theta}{dt}\right) = \left(\cfrac{1}{150}\right)(8)\)
\(\cfrac{1}{\sec^218^{\circ}}\cdot \sec^218^{\circ}\left(\cfrac{d\theta}{dt}\right) = \cfrac{1}{\sec^218^{\circ}} \cdot \cfrac{8}{150}\)
\(\cfrac{d\theta}{dt} = \cfrac{8}{150\sec^218^{\circ}}\)
\(\cfrac{d\theta}{dt} = \cfrac{8\cos^218^{\circ}}{150}\)
\(\cfrac{d\theta}{dt} = 0.048\; \left[\cfrac{\text{deg}}{\text{sec}}\right]\)
Finally, we can complete our table by filling in the missing values:
Variable |
Quantity |
Units |
Rate |
Rate Units |
Adjacent (a) |
\(a = 150\) |
\([\text{ft}]\) |
N/A |
N/A |
Opposite (b) |
\(b = 50\) |
\([\text{ft}]\) |
\(\cfrac{db}{dt} = 8\) |
\(\left[\cfrac{\text{ft}}{\text{sec}}\right]\) |
Angle (θ) |
\(θ = 18^{°}\) |
\([\text{deg}]\) |
\(\cfrac{dθ}{dt} = 0.048\) |
\(\left[\cfrac{\text{deg}}{\text{sec}}\right]\) |
Therefore, we can determine that the angle of elevation is increasing at at rate of \(0.048\;\left[\cfrac{\text{deg}}{\text{sec}}\right]\).
A particle \(P\) is moving along the graph of \(y = \sqrt{x^2-4}\), \(x \leq 2\) , so that the coordinate of \(P\) is increasing at the rate of \(5\) units per second. How fast is the \(y\)-coordinate of \(P\) increasing when \(x=3\)?
Show Answer
First, we can determine that \(x\) and \(y\) change in relation to time.
Their Related Rates are expressed as such:
\(x = \cfrac{dx}{dt}\)
\(y = \cfrac{dy}{dt}\)
Next, we can identify the rate at which the \(x\)-coordinate increases:
\(\cfrac{dx}{dt} = 5\; \left[\cfrac{\text{units}}{\text{second}}\right]\)
We can also determine what we need to find:
\(\cfrac{dy}{dt} = ?\) at \(x = 3\)
We can create a table listing all of the known values and what we still need to determine:
Variable |
Quantity |
Units |
Rate |
Rate Units |
\(x\)-coordinate (\(x\)) |
\(x = 3\) |
\([\text{units}]\) |
\(\cfrac{dx}{dt} = 5\) |
\(\left[\cfrac{\text{units}}{\text{sec}}\right]\) |
\(y\)-coordinate (\(y\)) |
\(y = \sqrt{x^2 - 4}\) |
\([\text{units}]\) |
\(\cfrac{dy}{dt} = ?\) |
\(\left[\cfrac{\text{units}}{\text{sec}}\right]\) |
Then, we can determine the value of \(y\) when \(x=3\):
\(y = \sqrt{(3)^2-4}\)
\(y = \sqrt{9-4}\)
\(y = \sqrt{5}\)
After, we can set the \(y\)-coordinate, \(\cfrac{dy}{dt}\), equal to its equation. We can then determine its derivative using Chain Rule and Implicit Differentiation:
\(\cfrac{dy}{dt} = (x^2-4)^{\frac{1}{2}}\)
\(\cfrac{dy}{dt} = \cfrac{1}{2}(x^2-4)^{-\frac{1}{2}} \left(2x\cdot \cfrac{dx}{dt}\right)\)
We can substitute \(x^2 - 4\) with a simpler expression:
\(y^2 = (\sqrt{x^2-4})^2\)
\(y^2 = x^2-4\)
We can now substitute all pertinent values and simplify to determine the speed of the \(y\)-coordinate:
\(\cfrac{dy}{dt} = \cfrac{1}{2}(y^2)^{-\frac{1}{2}} \left(2x\cdot \cfrac{dx}{dt}\right)\)
\(\cfrac{dy}{dt} = \cfrac{1}{2y} \left(2x\cdot \cfrac{dx}{dt}\right)\)
\(2y\cdot \cfrac{dy}{dt} = 2x\cdot \cfrac{dx}{dt}\)
\(2(\sqrt{5})\cdot \cfrac{dy}{dt} = 2(3)(5)\)
\(\cfrac{dy}{dt} = \cfrac{\cancel{2}(15)}{\cancel{2}(\sqrt{5})}\)
\(\cfrac{dy}{dt} = \cfrac{15}{\sqrt{5}} = 6.71\)
Finally, we can complete our table by filling in the missing values:
Variable |
Quantity |
Units |
Rate |
Rate Units |
\(x\)-coordinate (\(x\)) |
\(x = 3\) |
\([\text{units}]\) |
\(\cfrac{dx}{dt} = 5\) |
\(\left[\cfrac{\text{units}}{\text{sec}}\right]\) |
\(y\)-coordinate (\(y\)) |
\(y = \sqrt{5}\) |
\([\text{units}]\) |
\(\cfrac{dy}{dt} = 6.71\) |
\(\left[\cfrac{\text{units}}{\text{sec}}\right]\) |
Therefore, we can determine that the \(y\)-coordinate is moving at a speed of \(6.71\; \left[\cfrac{\text{units}}{\text{sec}}\right]\) when \(x=3\).