1. Applications of the Indefinite Integral
by M. Bourne
Displacement from Velocity, and Velocity from Acceleration
A very useful application of calculus is displacement, velocity and acceleration.
Recall (from Derivative as an Instantaneous Rate of Change) that we can find an expression for velocity by differentiating the expression for displacement:
Similarly, we can find the expression for the acceleration by differentiating the expression for velocity, and this is equivalent to finding the second derivative of the displacement:
It follows (since integration is the opposite process to differentiation) that to obtain the displacement, s of an object at time t (given the expression for velocity, v) we would use:
Similarly, the velocity of an object at time t, given the acceleration a, is given by:
Example 1:
A proton moves in an electric field such that its acceleration (in cms-2) is
a = -20(1+2t)-2, where t is in seconds.
Find the velocity as a function of time if v = 30 cms-1 when t = 0.
Here is how it is done using LiveMath:
Now for the normal answer:
- Answer
-
So
Put u = 1 + 2t then du = 2 dt
When t = 0, v = 30, so K = 20.
So the expression for velocity as a function of time is:
Example 2:
A flare is ejected vertically upwards from the ground at 15 m/s. Find the height of the flare after 2.5 s.
- Answer
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The object has acting on it the force due to gravity, so its acceleration is -9.8 ms-2.
Now at t = 0, the velocity = 15ms-1. So C = 15.
So the expression for velocity becomes:
Now, we need to find the displacement, so
Now, we know from the question that when t = 0, s = 0.
This gives K = 0.
So
At time t = 2.5, s = 6.875 m.
Displacement and Velocity Formulas
Using integration, we can obtain the well-known expressions for displacement and velocity, given a constant acceleration a, initial displacement zero, and an initial velocity v0:
v = ∫ a dt
v = at + K
Since the velocity at t = 0 is v0. we have K = v0. So:
v = v0 + at
Similarly,
s = ∫v dt = ∫(v0 + at) dt
s = v0t + at2/2 + C
Since the displacement at t = 0 is s = 0, we have C = 0. So:
Voltage across a Capacitor
Definition: The current, i (amperes), in an electric circuit equals the time rate of change of the charge q, (in coulombs) that passes a given point in the circuit. We can write this (with t in seconds) as:
By writing i dt = dq and integrating, we have:
The voltage, VC (in volts) across a capacitor with capacitance C (in farads) is given by
.
It follows that
You can see some more advanced applications of this at Applications of Ordinary Differential Equations.
Example 1:
The electric current (in mA) in a computer circuit as a function of time is i = 0.3 − 0.2t. What total charge passes a point in the circuit in 0.050s?
Here is the answer using LiveMath:
Normal answer:
- Answer
-
The charge, q, is given by:
At t = 0, q = 0, and this gives us K = 0.
So
(units are milli coulombs, as i was in mA.)
Example 2:
The voltage across an 8.50 nF capacitor in an FM receiver circuit is zero. Find the voltage after 2.00 μs if a current i = 0.042t (in mA) charges the capacitor.
- Answer
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nF = 10-9F; μs = 10-6s;
0.042t mA = 0.042 × 10-3t A
Now, we are told that when t = 0, VC = 0.
So K = 0.
Thus
So when t = 2.00 μs, we have:
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