Differential calculus is all about instantaneous rate of change. Let's see how this can be used to solve real-world word problems.
One way to interpret the derivative ff' of a function ff is that f(k)f'(k) is the instantaneous rate of change of ff at x=kx=k. Let's see how this interpretation can be used to solve word problems.
Say a water tank is being filled and the volume, in liters, of the water in the tank after tt seconds is given by the linear function V1(t)=23tV_1(t)=\dfrac23t.
The slope of the function, 23\dfrac23, represents its rate of change. In other words, the tank is being filled at a rate of 23\dfrac23 liters per second.
The rate of change of a linear function is always constant, which makes it relatively easy to reason about.
Now say a different tank is being filled, and this time the volume function V2(t)=0.1t2V_2(t)=0.1t^2 isn't linear.
Notice how the graph's growth is gradual at first and becomes more steep towards the end. The rate of change of V2V_2 isn't constant.
If we want to analyze the rate of change of V2V_2, we can talk about its instantaneous rate of change at any given point in time. The instantaneous rate of change of a function is given by the function's derivative.
V2(t)=0.2tV_2'(t)=0.2t
For example, V2(5)=1V_2'(5)=1. Mathematically, this means that the slope of the line tangent to the graph of V2V_2 when x=5x=5 is 11. What does this mean in the context of our water tank?
The tangent line's slope indicates the curve's slope at that particular point in time. Since we already saw how slope gives us the rate of change, we can interpret V2(5)=1V_2'(5)=1 as follows:
At t=5t=5 seconds, the tank is being filled at a rate of 11 liter per second.
Notice a couple things about this interpretation:
First, the rate is given in liters per second. The units of a derivative are always a ratio of the dependent quantity (e.g. liters) over the independent quantity (e.g. seconds).
Second, the rate is given for a specific point in time (e.g. t=5t=5 seconds). This is because it's instantaneous. Take another point in time, and the rate might be different. Look at an interval of time, and the rate isn't constant.
Problem 1.A
In Problem set 1 we will analyze the following context:
Lindsay is walking home from school. Her distance from school, in meters, after tt minutes is modeled by the differentiable function DD.
What units should we use to measure D(t)D'(t)?
Choose 1 answer:
Choose 1 answer:

Problem 2
HH gives the height of a tree, in centimeters, tt weeks after it was planted.
Four students were asked to interpret the meaning of H(5)=3H'(5)=3 in this context.
Can you match the teacher's comments to the interpretations?
Interpretation
Teacher's comments
  • At 55 weeks, the rate at which the tree is growing is 33.
  • At 55 weeks, the rate at which the tree is growing is 33 centimeters per week.
  • The rate at which the tree is growing over the first 55 weeks is 33 centimeters per week.
  • At 55 weeks the tree was 33 centimeters tall.
  • Notice that we are talking about HH', not HH.
  • Kudos! You are correct.
  • You didn't use appropriate units.
  • Recall that the derivative is the instantaneous rate of change.

Common mistake: Forgetting to include units, or using incorrect units

Remember: When analyzing problems in applied contexts, remember we should always use units.
For example, in Problem 2, HH gets an input that is measured in weeks and gives an output that is measured in centimeters. Its derivative HH' also gets an input that is measured in weeks, but its output is the rate centimeters per week.

Another common mistake: Using phrases that refer to “over an interval of time” rather than “at a point in time”

Derivatives are all about instantaneous rate of change. Therefore, when we interpret the rate of a function given the value of its derivative, we should always refer to the specific point when that rate applies.

Solving problems that involve instantaneous rate of change

Consider the following problem:
Carlos has taken an initial dose of a prescription medication. The amount of medication, in milligrams, in Carlos's bloodstream after tt hours is given by the following function:
M(t)=20e0.8tM(t)=20\cdot e^{^{\large -0.8t}}
What is the instantaneous rate of change of the remaining amount of medication after 11 hour?
The first thing that should pop up to us as we read this problem is that we are asked for the instantaneous rate of change of a quantity. This means we need to use derivatives.
The only function whose derivative we can use is MM, but let's make sure this is what we want: MM gives the amount of medication in Carlos's bloodstream over time, and we are asked for the instantaneous rate of change of that amount. So yes, we need MM':
M(t)=16e0.8tM'(t)=-16\cdot e^{-0.8t}
We are asked for the instantaneous rate of change after 11 hour, which means we need to evaluate MM' at t=1t=1:
M(1)=16e0.87.2M'(1)=-16\cdot e^{-0.8}\approx -7.2
Finally, we need to remember to use units. Since MM gives an amount in milligrams for a given input in hours, the units in which we measure MM' are milligrams per hours.
In conclusion, the instantaneous rate of change of the remaining amount of medication after 11 hour is 7.2-7.2 milligrams per hour.
Problem 3
CC gives the cost, in dollars, to shred ww pounds of confidential documents of a company.
C(w)=0.001w30.15w2+7.5wC(w)=0.001w^3-0.15w^2+7.5w
What is the instantaneous rate of change of the costs when the weight of the documents is 1010 pounds?
Choose 1 answer:
Choose 1 answer:

Want more practice? Try this exercise.

Common mistake: Evaluating the original function rather than the derivative

Remember: When we are asked about the rate of change of a function ff, we want to look for the derivative ff'. Evaluating ff at a point will not provide us any information about the rate of change of ff at that point.
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