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Common applications of the derivative
Derivatives allow us to find instantaneous rates of change. Here are some common applications of derivatives.
Position-velocity-acceleration
This is the example demonstrated in the graph at the top of this page. The position function, p(x), is graphed in red. The velocity function, v(x), is in blue, and the acceleration, a(x), is in green. Recall, when differentiating using the Power Rule, you subtract 1 from the exponent's power. Notice in the equations how the position, velocity, and acceleration functions all decrease in degree by 1? The derivative of the position function is the velocity function; the derivative of the velocity function (the second derivative of the position function) is the acceleration function! Some of my students have referred to this as the P-V-A relationship.
One important concept to understand is that the functions give instantaneous values, not average values (these are derivatives, which are instantaneous rates of change.) For example, y-values on the velocity function illustrate instantaneous velocities at each x-value of time. Also, recall the notation of the derivative dy/dx. The units of each of these function values will follow the "division" behavior and each subsequent derivative's units will be divided by another dimension of time. For example, if the position of an object is measured in feet during a course of time measured in seconds, the velocity of that object would be measured in feet/second; the acceleration of that object could then be measured in feet/second/second or feet/second^2.
Practice Problems:
One important concept to understand is that the functions give instantaneous values, not average values (these are derivatives, which are instantaneous rates of change.) For example, y-values on the velocity function illustrate instantaneous velocities at each x-value of time. Also, recall the notation of the derivative dy/dx. The units of each of these function values will follow the "division" behavior and each subsequent derivative's units will be divided by another dimension of time. For example, if the position of an object is measured in feet during a course of time measured in seconds, the velocity of that object would be measured in feet/second; the acceleration of that object could then be measured in feet/second/second or feet/second^2.
Practice Problems:
![Picture](/uploads/2/6/0/9/26090300/9070994.png?678)
Here are the answers to the above Position-Velocity-Acceleration problems:
1) B
2) a(3) = -1.577
3) D
4) a. v(t) = 3t^2 - 1
b. v(3) = 26
Related rates
Another common application of the derivative is the problem of related rates. When given a function, not all the variables will change at the same rate, or even proportionally to each other. For example, just because the the radius of a balloon is increasing at a constant rate of 1"/sec does not mean the volume or surface area of the balloon is increasing at that rate. (In fact, they can't have the same rate of change, the units of inches per second are incorrect for volume or surface area.) Here are some general steps to follow when solving related rates problems:
Practice Problems:
- Write the primary equation, involving the variable whose rate you need to find.
- Write any secondary equations necessary. (You especially might need to do this if your primary equation has more than two variables. If your primary equation has variables for whom you do not know a derivative expression or relationship, you will need to express these variables in terms of the other variables prior to differentiating.)
- Differentiate the (adjusted, if necessary) primary equation with respect to time. Don't forget to implicitly differentiate.
- Substitute in all initial conditions and solve for the unknown rate.
Practice Problems:
![Picture](/uploads/2/6/0/9/26090300/9791707.png?665)
Here are the answers to the above Related Rates problems:
1) 0.071 or 0.072 in/min
2) a. V = 5.5*b*h
b. dh/dt = -5/66 or 0.075 or 0.076 ft/min
3) a. dV/dt = 63pi or 197.928 cubic cm/sec
b. dA/dt = 0.42pi or 1.319 square cm/sec (Hint: First find r when V = 36pi.)
c. r = 0.282 cm
1) 0.071 or 0.072 in/min
2) a. V = 5.5*b*h
b. dh/dt = -5/66 or 0.075 or 0.076 ft/min
3) a. dV/dt = 63pi or 197.928 cubic cm/sec
b. dA/dt = 0.42pi or 1.319 square cm/sec (Hint: First find r when V = 36pi.)
c. r = 0.282 cm