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11.5. Physical Applications of Derivatives

With the widespread use of calculus in all areas of science, especially physics, for which Isaac Newton invented it to begin with, it would be a shame to omit a discussion of applications of differential calculus to physics. Thus, we close this chapter with a brief look at how derivatives are used in physics.

Many physical quantities are derivatives of other quantities. One of the most basic examples is velocity (speed and direction), which is the derivative of position. For our discussion, we will only consider motion in one direction, although in general, motion may occur simultaneously in three independent directions, corresponding to the three dimensions of space. Let x(t) denote the position of an object at time t. Typically, x is measured in meters and time in seconds. Then the velocity v(t) of the object is equal to the derivative of its position. In other words, we have

**(11.5.1)**v(t) = dx / dt.

Furthermore, the * acceleration* a(t) of the object is defined as the derivative of its velocity, i.e. we have

**(11.5.2)**a(t) = dv / dt.

**Example 1: **The position of an object at time t is given by the equation x(t) = 3t^{2} - 8t + 5. Compute the position, velocity and acceleration of the object at t=0, t=1, and t=2.

**Solution: **We have v(t) = dx / dt = 6t - 8 and a(t) = dv / dt = 6. Thus we have

x(0) = 5; v(0) = -8; a(0) = 6

x(1) = 0; v(1) = -2; a(1) = 6

x(2) = 1; v(2) = 4; a(2) = 6

Let us now return to our discussion from Section 1.10 of the law of falling bodies. We mentioned that if we drop an object from rest, the distance it falls after t seconds is equal to y(t) = 4.9t^{2} and its speed after t seconds is given by v(t) = 9.8t. We now want to generalize this law to the case of an object which is thrown directly up or down from an arbitrary height above the ground. We denote the height from which the object is thrown by y_{0} and the velocity at which it is thrown by v_{0}, which is positive if the object is thrown upward and negative if it is thrown downward. Then the formula for the height of the object above the ground after t seconds is given by

**(11.5.3)**y(t) = y_{0}+ v_{0}t - 4.9 t^{2}

where as always, height is measured in meters and time in seconds. Taking the derivative of Equation (11.5.3), we find the velocity of the object is given by

**(11.5.4)**v(t) = dy / dt = v_{0}- 9.8 t.

Taking the derivative once again, we get the acceleration of the object, which is given by

**(11.5.5)**a(t) = dv / dt = -9.8.

Note that the acceleration of the object is constant; it is always equal to -9.8 meters per second per second. The fact that it is negative means the object is accelerating toward the ground. The rate at which object accelerate toward the ground during free-fall is a physical constant, which goes by the symbol g, which stands for the acceleration due to gravity. Thus, we have g = 9.8 meters per second per second. (Actually, g is not exactly constant, but varies slightly with latitude and altitude, but the maximum variation along the earth's surface is only about 1 part in 200.)

**Example 2: **A ball is thrown upward from a height of 10 meters at a speed of 30 meters per second. Determine the height of the ball above the ground after 3 seconds and after 6 seconds.

**Solution: **We have y_{0} = 10 and v_{0} = 30. Plugging in these values into (11.5.3), we find the equation for the height of the ball after t seconds is given by

y(t) = 10 + 30 t - 4.9 t^{2}.

Plugging in t = 3, we find the height of the ball after 3 seconds is given by

y(3) = 10 + (30)(3) - (4.9)(3^{2}) = 10 + 90 - 44.1 = 55.9 meters.

Plugging in t = 6, we find the height of the ball after 6 seconds is given by

y(3) = 10 + (30)(6) - (4.9)(6^{2}) = 10 + 180 - 176.4 = 13.6 meters.

**Example 3: **Determine the velocity of the ball in Example 2 after 3 seconds and after 6 seconds.

**Solution: **The velocity after t seconds is given by Equation (11.5.4), with v_{0} = 30. Thus we have

v(t) = 30 - 9.8 t.

Plugging in t = 3, we find the velocity of the ball after 3 seconds is given by

v(3) = 30 - (9.8)(3) = 30 - 29.4 = 0.6,

which means the ball is moving upwards at 0.6 meters per second. The velocity of the ball after 6 seconds is given by

v(6) = 30 - (9.8)(6) = 30 - 58.8 = -28.8,

which means the ball is now moving downwards at 28.8 meters per second.