Find an equation of the tangent line to the graph of a function at a point.
Using Instantaneous Rates of Change to Solve Real-World Problems
Another way to interpret an instantaneous rate of change at [latex]x=a[/latex] is to observe the function in a real-world context. The unit for the derivative of a function [latex]f\left(x\right)[/latex] is
[latex]\frac{\text{output units }}{\text{ input unit }}[/latex]
Such a unit shows by how many units the output changes for each one-unit change of input. The instantaneous rate of change at a given instant shows the same thing: the units of change of output per one-unit change of input.
One example of an instantaneous rate of change is a marginal cost. For example, suppose the production cost for a company to produce [latex]x[/latex] items is given by [latex]C\left(x\right)[/latex], in thousands of dollars. The derivative function tells us how the cost is changing for any value of [latex]x[/latex] in the domain of the function. In other words, [latex]\begin{align}{C}^{\prime }\left(x\right)\end{align}[/latex] is interpreted as a marginal cost, the additional cost in thousands of dollars of producing one more item when [latex]x[/latex] items have been produced. For example, [latex]\begin{align}{C}^{\prime }\left(11\right)\end{align}[/latex] is the approximate additional cost in thousands of dollars of producing the 12th item after 11 items have been produced. [latex]\begin{align}{C}^{\prime }\left(11\right)=2.50\end{align}[/latex] means that when 11 items have been produced, producing the 12th item would increase the total cost by approximately $2,500.00.
The cost in dollars of producing [latex]x[/latex] laptop computers in dollars is [latex]f\left(x\right)={x}^{2}-100x[/latex]. At the point where 200 computers have been produced, what is the approximate cost of producing the 201st unit?
If [latex]f\left(x\right)={x}^{2}-100x[/latex] describes the cost of producing [latex]x[/latex] computers, [latex]\begin{align}{f}^{\prime }\left(x\right)\end{align}[/latex] will describe the marginal cost. We need to find the derivative. For purposes of calculating the derivative, we can use the following functions:
[latex]\begin{align}\text{ } \\ {f}^{\prime }\left(x\right)&=\frac{f\left(a+h\right)-f\left(a\right)}{h}&& \text{Formula for a derivative} \\ &=\frac{{\left(x+h\right)}^{2}-100\left(x+h\right)-\left({x}^{2}-100x\right)}{h}&& \text{Substitute }f\left(a+h\right)\text{ and }f\left(a\right). \\ &=\frac{{x}^{2}+2xh+{h}^{2}-100x - 100h-{x}^{2}+100x}{h}&& \text{Multiply polynomials, distribute}. \\ &=\frac{2xh+{h}^{2}-100h}{h}&& \text{Collect like terms}. \\ &=\frac{\cancel{h}\left(2x+h - 100\right)}{\cancel{h}}&& \text{Factor and cancel like terms}. \\ &=2x+h - 100&& \text{Simplify}. \\ &=2x - 100&& \text{Evaluate when }h=0. \\ {f}^{\prime }\left(x\right)&=2x - 100 && \text{Formula for marginal cost} \\ {f}^{\prime }\left(200\right)&=2\left(200\right)-100=300&& \text{Evaluate for 200 units}. \end{align}[/latex]
The marginal cost of producing the 201st unit will be approximately $300.
A car leaves an intersection. The distance it travels in miles is given by the function [latex]f\left(t\right)[/latex], where [latex]t[/latex] represents hours. Explain the following notations:
First we need to evaluate the function [latex]f\left(t\right)[/latex] and the derivative of the function [latex]\begin{align}{f}^{\prime }\left(t\right)\end{align}[/latex], and distinguish between the two. When we evaluate the function [latex]f\left(t\right)[/latex], we are finding the distance the car has traveled in [latex]t[/latex] hours. When we evaluate the derivative [latex]\begin{align}{f}^{\prime }\left(t\right)\end{align}[/latex], we are finding the speed of the car after [latex]t[/latex] hours.
[latex]f\left(0\right)=0[/latex] means that in zero hours, the car has traveled zero miles.
[latex]\begin{align}{f}^{\prime }\left(1\right)=60\end{align}[/latex] means that one hour into the trip, the car is traveling 60 miles per hour.
[latex]f\left(1\right)=70[/latex] means that one hour into the trip, the car has traveled 70 miles. At some point during the first hour, then, the car must have been traveling faster than it was at the 1-hour mark.
[latex]f\left(2.5\right)=150[/latex] means that two hours and thirty minutes into the trip, the car has traveled 150 miles.
Finding the Instantaneous Speed of a Particle
If a function measures position versus time, the derivative measures displacement versus time, or the speed of the object. A change in speed or direction relative to a change in time is known as velocity. The velocity at a given instant is known as instantaneous velocity.
In trying to find the speed or velocity of an object at a given instant, we seem to encounter a contradiction. We normally define speed as the distance traveled divided by the elapsed time. But in an instant, no distance is traveled, and no time elapses. How will we divide zero by zero? The use of a derivative solves this problem. A derivative allows us to say that even while the object’s velocity is constantly changing, it has a certain velocity at a given instant. That means that if the object traveled at that exact velocity for a unit of time, it would travel the specified distance.
instantaneous velocity
Let the function [latex]s\left(t\right)[/latex] represent the position of an object at time [latex]t[/latex]. The instantaneous velocity or velocity of the object at time [latex]t=a[/latex] is given by
A ball is tossed upward from a height of 200 feet with an initial velocity of 36 ft/sec. If the height of the ball in feet after [latex]t[/latex] seconds is given by [latex]s\left(t\right)=-16{t}^{2}+36t+200[/latex], find the instantaneous velocity of the ball at [latex]t=2[/latex].
First, we must find the derivative [latex]\begin{align}{s}^{\prime }\left(t\right)\end{align}[/latex] . Then we evaluate the derivative at [latex]t=2[/latex], using [latex]s\left(a+h\right)=-16{\left(a+h\right)}^{2}+36\left(a+h\right)+200[/latex] and [latex]s\left(a\right)=-16{a}^{2}+36a+200[/latex].