Related rates are a fundamental concept in calculus that involve finding the rate at which one quantity changes with respect to another. This often comes in handy in real-world scenarios, like tracking the growth of an object or the flow of a liquid. In this guide, we’ll explore three practical examples that illustrate how related rates work. Each example will provide context, a detailed problem, and relevant notes. Let’s dive in!
Imagine you have a conical tank that is being filled with water. The radius of the base of the cone is 2 feet, and the height is 6 feet. Water is being poured into the tank at a constant rate of 3 cubic feet per minute. We want to find out how fast the water level is rising when the water is 4 feet deep.
To solve this, we first need to establish a relationship between the volume of the cone and the height of the water. The formula for the volume of a cone is:
[ V = \frac{1}{3} \pi r^2 h ]
In our case, the radius and height are related because they form a similar triangle. At the height of 6 feet, the radius is 2 feet, so:
[ \frac{r}{h} = \frac{2}{6} \Rightarrow r = \frac{1}{3}h ]
Substituting this back into the volume formula gives us:
[ V = \frac{1}{3} \pi \left(\frac{1}{3}h\right)^2 h = \frac{\pi}{27}h^3 ]
Now, taking the derivative with respect to time (t):
[ \frac{dV}{dt} = \frac{\pi}{9}h^2 \frac{dh}{dt} ]
We know ( \frac{dV}{dt} = 3 ) cubic feet per minute. Plugging in the height of 4 feet gives:
[ 3 = \frac{\pi}{9}(4^2) \frac{dh}{dt} ]
Solving for ( \frac{dh}{dt} ) yields:
[ \frac{dh}{dt} = \frac{3 \cdot 9}{16\pi} \approx 0.17 \text{ feet per minute} ]
This means the water level is rising at approximately 0.17 feet per minute when the water is 4 feet deep.
Consider a 10-foot ladder leaning against a vertical wall. The bottom of the ladder is sliding away from the wall at a rate of 1 foot per second. We want to find out how fast the top of the ladder is sliding down the wall when the bottom is 6 feet away from the wall.
Let’s define our variables:
Using the Pythagorean theorem, we know:
[ x^2 + y^2 = 10^2 ]
Taking the derivative with respect to time gives us:
[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 ]
Now, we need to find the height y when x = 6.
[ 6^2 + y^2 = 100 \Rightarrow y^2 = 64 \Rightarrow y = 8 ]
Now we can substitute into the derivative equation:
[ 2(6)(1) + 2(8) \frac{dy}{dt} = 0 ]
Solving for ( \frac{dy}{dt} ) gives:
[ 12 + 16 \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{12}{16} = -0.75 \text{ feet per second} ]
This means the top of the ladder is sliding down the wall at a rate of 0.75 feet per second when the bottom is 6 feet from the wall.
Let’s say you have a spherical balloon that is being inflated, and its radius increases at a constant rate of 0.5 inches per minute. We want to find the rate at which the volume of the balloon is increasing when the radius is 3 inches.
The formula for the volume of a sphere is:
[ V = \frac{4}{3} \pi r^3 ]
Differentiating with respect to time, we get:
[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} ]
Now, we know ( \frac{dr}{dt} = 0.5 ) inches per minute and want to find ( \frac{dV}{dt} ) when r = 3 inches:
[ \frac{dV}{dt} = 4\pi (3^2)(0.5) = 4\pi (9)(0.5) = 18\pi \approx 56.55 \text{ cubic inches per minute} ]
Thus, the volume of the balloon is increasing at approximately 56.55 cubic inches per minute when the radius is 3 inches.
These examples provide a solid foundation for understanding related rates in calculus. By applying these concepts to real-world situations, you can see how calculus helps us make sense of change in our daily lives.