By utilizing the concept of the rate of change, we can apply derivatives to tackle real-world scenarios, especially in understanding how one quantity changes in relation to another. This approach is known as related rates in AP Calculus AB.
Notice how the radius and its rate of change are clearly labeled in the above illustration. Labeling your quantities in relation to the image is an excellent strategy to identify the known and unknown quantities of the problem. Here, we observe that the known quantities are a radius of 30 meters and a rate of change of the radius with respect to time, denoted as
m/s is 3 meters per second (m/s).
Since we need to find the area at a specific radius and its rate of change, we can model the relationship using the area formula for a circle:
Given that the problem asks for how fast the area increases at a particular radius (when ( r = 30)), we can express this as “the rate of change of area with respect to time,” denoted .
Now that we know what quantity we are seeking, we can differentiate the formula with respect to time. Using implicit differentiation, we find:
This can be simplified to:
At this point, you may notice we have most of the quantities present in this derivative, allowing us to substitute our known values and solve for the unknown quantity. Be sure to include the units!
Thus, our final answer is 180πm2 /s
Here is another problem:
Suppose a 15 ft tall ladder is leaning against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 5 ft/s, how quickly is the top of the ladder sliding down when the bottom of the ladder is 9 ft away from the wall?
At first glance, this problem may seem less straightforward than the previous one. This is a classic "ladder" problem. To grasp what we need to find, it’s helpful to draw or model the situation, using the quantities provided:
Here, we visualize the ladder leaning against the vertical y-axis, while the bottom of the ladder reaches the horizontal x-axis. This forms a right triangle, and since we know the length of the ladder (the hypotenuse) and one side's length, we can apply the Pythagorean Theorem to model the relationship between the lengths of the wall, the ground, and the ladder.
x2 + y2 = 225
Now that we have a formula that models our scenario, we see that our unknown quantity is the rate of change in (y) because as the ladder slides away from the wall (increasing the x-value), the height of the ladder against the wall decreases. This is denoted as . We can use implicit differentiation on the formula to find the rates of change for all quantities with respect to time:
At this point, you might realize that while we have values for most quantities in this derivative, we still need to find (y). You may feel confused, but all you need to do is calculate (y) using the original formula.
Given that (x = 9) ft, we can find the value of (y) by:
Now that we have the value of (y), we can return to our derivative and substitute the known values to solve for :
By substituting the known quantities into the equation, we can use algebra to solve for , which
gives us the answer!
Key Terms to Review
26 videos|75 docs|38 tests
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1. What are related rates in calculus? | ![]() |
2. How do you approach a related rates problem? | ![]() |
3. Can you give an example of a related rates problem? | ![]() |
4. What is the significance of the Pythagorean theorem in related rates problems? | ![]() |
5. How do I differentiate an area formula with respect to time? | ![]() |