Lesson 14: Related Rates
Related Rates (Part I)
Example 1
Assume \(x\) and \(y\) are both differentiable functions of \(t\) and that \(8x^5y = 16\). Find \(\frac{dy}{dt}\) if \(\frac{dx}{dt} = 8\) and \(x = 1\).
Solution:
We use implicit differentiation and take the derivative of both sides of the equation with respect to \(t\):
\[\frac{d}{dt}(8x^5y) = \frac{d}{dt}(16)\]Applying the product rule (and chain rule too) on the left side:
\[40x^4 \cdot \frac{dx}{dt} \cdot y + 8x^5 \cdot \frac{dy}{dt} = 0\]Note that if \(x = 1\), then from the original equation, \(8\cdot 1^5\cdot y = 16\), and so \(y = 2\).
Substituting \(\frac{dx}{dt} = 8\), \(x = 1\), and \(y = 2\):
\[40\cdot 1^4 \cdot 8 \cdot 2 + 8\cdot 1^5 \cdot \frac{dy}{dt} = 0 \implies 640 + 8\frac{dy}{dt} = 0 \implies 8\frac{dy}{dt} = -640 \implies \frac{dy}{dt} = -80\]General Word Problem Approach
- Read the problem, and identify any numerical values.
- Draw a picture to represent the problem.
- Write down an equation which describes the picture in step 2. Each variable in this equation is a function of time. Time is given by the variable \(t\).
- Use implicit differentiation to differentiate both sides of the equation in step 3 with respect to time (i.e., \(t\)).
- Substitute the given information into the equation from step 4 and solve for what we want to find.
Examples
Example 2
The radius of a circle is increasing at the rate of 5 ft/min.
(a) Find the rate of change of the circumference of the circle when \(r = 6\) ft.
Solution:
Diagram Description: A circle with radius \(r = 6\) ft marked from the center to the edge.
The circumference of a circle is given by \(C = 2\pi r\), where both \(C\) and \(r\) are functions of time.
Using implicit differentiation, taking the derivative with respect to \(t\) of both sides:
\[\frac{d}{dt}(C) = \frac{d}{dt}(2\pi r)\implies\frac{dC}{dt} = 2\pi \cdot \frac{dr}{dt}\]We are given that \(\frac{dr}{dt} = 5\) ft/min. Substituting:
\[\frac{dC}{dt} = 2\pi\cdot 5 = 10\pi\]Therefore, the rate of change of the circumference is \(10\pi\) ft/min, or approximately 31.42 ft/min.
(b) Find the rate of change of the area of the circle when \(r = 6\) ft.
Solution:
Diagram Description: A circle with radius \(r = 6\) ft marked from the center to the edge.
The area of a circle is given by \(A = \pi r^2\), where both \(A\) and \(r\) are functions of time.
Using implicit differentiation, taking the derivative with respect to \(t\) of both sides:
\[\frac{d}{dt}(A) = \frac{d}{dt}(\pi r^2) \implies \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}\]We are given that \(\frac{dr}{dt} = 5\) ft/min and \(r = 6\) ft. Substituting:
\[\frac{dA}{dt} = 2\pi\cdot 6\cdot 5 = 60\pi\]Therefore, the rate of change of the area is \(60\pi\) ft²/min, or approximately 188.50 ft²/min.
Example 3
The length of a rectangle is decreasing at a rate of 3 in/s, and its width is decreasing at a rate of 2 in/s. When the length is 10 inches and the width is 8 inches, how fast is the area of the rectangle decreasing?
Solution:
Diagram Description: A rectangle with area \(A\), width \(w = 8\) in, and length \(\ell = 10\) in.
The area of a rectangle is given by \(A = \ell \cdot w\), where both \(\ell\) and \(w\) are functions of time.
Using implicit differentiation, taking the derivative with respect to \(t\) of both sides:
\[\frac{d}{dt}(A) = \frac{d}{dt}(\ell \cdot w)\]Applying the product rule:
\[\frac{dA}{dt} = \frac{d\ell}{dt} \cdot w + \ell \cdot \frac{dw}{dt}\]We are given that \(\frac{d\ell}{dt} = -3\) in/s (negative because decreasing), \(\frac{dw}{dt} = -2\) in/s, \(\ell = 10\) in, and \(w = 8\) in. Substituting:
\[\frac{dA}{dt} = (-3)\cdot 8 + 10\cdot(-2)\implies \frac{dA}{dt} = -24 - 20 = -44\]Therefore, the area of the rectangle is decreasing at a rate of 44 in²/s.
Example 4
The radius of a spherical balloon is decreasing at a rate of 1.9 cm/s. How fast is the volume decreasing when the radius is 16 cm? Note that the volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\), where \(r\) is the radius of the sphere.
Solution:
Diagram Description: A sphere with radius \(r = 16\) cm, shown with a dashed line from the center to the edge.
We are given \(V = \frac{4}{3}\pi r^3\), where both \(V\) and \(r\) are functions of time.
Using implicit differentiation, taking the derivative with respect to \(t\) of both sides:
\[\frac{d}{dt}(V) = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) \implies \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\]We are given that \(\frac{dr}{dt} = -1.9\) cm/s (negative because decreasing) and \(r = 16\) cm. Substituting:
\[\frac{dV}{dt} = 4\pi\cdot 16^2\cdot (-1.9)\implies \frac{dV}{dt} = 4\pi\cdot 256\cdot (-1.9)\implies \frac{dV}{dt} = -1945.6\pi\]Therefore, the volume of the sphere is decreasing at a rate of \(1945.6\pi\) cm³/s, or approximately 6112.28 cm³/s.
Example 5
Gravel is dumped out of a truck onto the ground at 13 cm³/s, forming a conical pile with a base diameter that is always equal to the pile's altitude (i.e., height). How fast is the altitude of the pile increasing when the pile is 9 cm high? Note that the volume of a cone is \(\frac{1}{3}\pi r^2h\), where \(r\) is the radius of the base and \(h\) is the height of the cone.
Solution:
Diagram Description: A cone with height \(h = 9\) cm and base radius \(r\). The relationship \(2r = h\) is shown, which means \(r = \frac{h}{2}\). The height is measured from the apex (top) to the base.
We are given \(V = \frac{1}{3}\pi r^2h\). Also, the base diameter equals the height, and so \(2r = h\), implying \(r = \frac{h}{2}\). Substituting \(r = \frac{h}{2}\) into the volume formula:
\[V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi h^3}{12}\]Using implicit differentiation, taking the derivative with respect to \(t\) of both sides:
\[\frac{d}{dt}(V) = \frac{d}{dt}\left(\frac{\pi h^3}{12}\right)\implies \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}\]We are given that \(\frac{dV}{dt} = 13\) cm³/s (positive because volume is increasing) and \(h = 9\) cm. Substituting:
\[13 = \frac{\pi\cdot 9^2}{4} \cdot \frac{dh}{dt}\implies 13 = \frac{81\pi}{4} \cdot \frac{dh}{dt}\implies \frac{dh}{dt} = \frac{13 \cdot 4}{81\pi} = \frac{52}{81\pi}\]Therefore, the altitude of the pile is increasing at a rate of \(\frac{52}{81\pi}\) cm/s, or approximately 0.204 cm/s.