Video Transcript
In this video, we’re going to learn
how to apply integrals to solve problems involving motion in a straight line. We’ll begin by recapping the
methods for integrating some key functions and the process of integration by
substitution before considering how integration links to motion along a straight
line. We’ll then consider a variety of
examples that demonstrate these techniques.
The calculus techniques we’ll be
using in this video are the following. We need to know how to integrate a
power of 𝑥. That’s a function of the form 𝑎𝑥
to the 𝑛th power, where 𝑎 and 𝑛 are real constants and 𝑛 is not equal to
negative one. We add one to the power and divide
by that new value. So the integral of 𝑎𝑥 to the 𝑛th
power is 𝑎 𝑥 to the 𝑛 plus one power divided by 𝑛 plus one, plus of course we
have that constant of integration, 𝑐. In the case where 𝑛 is equal to
negative one, we’re integrating a function of the form 𝑎 over 𝑥, which gives us a
result of 𝑎 times the natural log of the absolute value of 𝑥 plus 𝑐. We’ll also be considering
exponential and trigonometric functions whose integrals are as shown. Finally, it’s beneficial that
you’re confident in applying integration by substitution, though you will be able to
access the majority of this video without this skill.
So how does calculus link to
motion? In particular, rectilinear
motion. That’s motion along a straight
line. Let’s recall the definitions for
displacement, velocity, and acceleration. Displacement — sometimes given as
𝑠 in function notation — is the vector that describes the position of an object
away from a given starting point. Velocity is then the rate of change
of the displacement of this object with respect to time. And acceleration is the rate of
change of the velocity of the object with respect to time. And this means if we define 𝑠 to
be a function that measures displacement at time 𝑡, then the velocity would be the
derivative of 𝑠 with respect to 𝑡. In function notation, that’s 𝑠
prime of 𝑡. And in Leibniz’s notation, you
might see it as d𝑠 by d𝑡.
Similarly, we can say that if 𝑣 is
a function that measures velocity at time 𝑡, then the acceleration is the
derivative of 𝑣 with respect to 𝑡. It’s 𝑣 prime of 𝑡 or d𝑣 by
d𝑡. And of course, since 𝑣 is the
derivative of 𝑠 with respect to 𝑡, then we can say that acceleration can also be
written as the second derivative of 𝑠. That’s 𝑠 double prime of 𝑡. And since integration is the
opposite process of differentiation, we can draw a little flow diagram to show the
relationship between displacement, velocity, and acceleration. To find an expression for velocity,
given an expression for acceleration in terms of time, we integrate. Similarly, to find an expression
for displacement, given an expression for velocity in terms of time, we integrate
with respect to time. Now it is also worth remembering
that distance and speed are scalar quantities, sometimes called the magnitude of
displacement and velocity, respectively.
Let’s now have a look at a number
of examples that demonstrate these ideas.
A particle is moving in a straight
line such that its velocity at time 𝑡 seconds is given by 𝑣 equals 15𝑡 squared
minus eight 𝑡 meters per second when 𝑡 is greater than or equal to zero. Given that its initial position
from a fixed point is 20 meters, find an expression for its displacement at time 𝑡
seconds.
Remember, velocity is the rate of
change of displacement of an object. This means we can differentiate a
function for displacement to find a function for velocity. Conversely, we can say that the
integral of the function in velocity will provide us with a function for
displacement. So to answer this question, we’re
going to integrate our function with respect to 𝑡 and then use the information
about the initial position to find a full expression for the displacement. 𝑠 is equal to the integral of 15𝑡
squared minus eight 𝑡 with respect to 𝑡.
Remember, the integral of the sum
or difference of two or more functions is equal to the sum or difference of the
integral of each function. So we can actually integrate 15𝑡
squared and negative eight 𝑡 individually. We know also that to integrate a
term of this form, we simply add one to the power and then divide by this new
number. And we obtain the interval of 15𝑡
squared to be 15𝑡 cubed divided by three. Now, in fact, we obtain a constant
of integration too. But we’ll deal with that in a
moment.
The integral of negative eight 𝑡
is negative eight 𝑡 squared divided by two. Then, we combine the two constants
of integration obtained by integrating 15𝑡 squared and negative eight 𝑡. And we see that 𝑠 is equal to 15𝑡
cubed over three plus negative eight 𝑡 squared over two plus 𝑐. We can simplify this somewhat. And what we’ve actually obtained is
the general equation for the displacement of our object. It’s five 𝑡 cubed minus four 𝑡
squared plus 𝑐. Now we can actually work out the
particular equation. That involves finding the value of
𝑐. And we can do that because we’re
told the initial position from a fixed point, in other words, its initial
displacement. We’re told its initial position
from the fixed point is 20 meters. So when 𝑡 is equal to zero, 𝑠 is
equal to 20.
So let’s substitute these values
into our result. We have 20 equals five times zero
cubed minus four times zero cubed plus 𝑐. Well, that just simplifies to 20
equals 𝑐. So the answer here is 𝑠 equals
five 𝑡 cubed minus four 𝑡 squared plus 20 meters. And to remember of course, we can
reverse this process and differentiate this expression with respect to 𝑡 to check
our solution. When we do, we obtain d𝑠 by
d𝑡. And that’s of course read to be
three times five 𝑡 squared minus two times four 𝑡, which simplifies to 15𝑡
squared minus eight 𝑡 as required.
We’ll now see how we can use
integration twice to solve problems involving rectilinear motion.
A particle is moving in a straight
line such that its acceleration at time 𝑡 seconds is given by 𝑎 equals two 𝑡
minus 18 meters per seconds squared, where 𝑡 is greater than or equal to zero. Given that its initial velocity is
20 meters per second and its initial displacement is zero meters, find an expression
for the displacement of the particle at time 𝑡.
Remember, acceleration is the rate
of change of the velocity of an object. This means we differentiate the
function for velocity to find the function for acceleration. We can conversely say that the
integral of the function for acceleration will provide us with the function for
velocity. Similarly, velocity is equal to the
derivative of displacement with respect to time. So we can say that to find
displacement, we can integrate the function for velocity with respect to time.
To answer this question then, we’re
going to need to integrate our function for acceleration with respect to time
twice. Throughout this process, we’ll use
the fact that we know its initial velocity and its initial displacement. This will help us to find a
particular expression for displacement. So velocity is the indefinite
integral of two 𝑡 minus 18 with respect to 𝑡. The integral of two 𝑡 is two 𝑡
squared over two. The integral of negative 18 is
negative 18𝑡. And of course, we mustn’t forget
that we have this constant of integration 𝑐. And we see that 𝑣 is equal to 𝑡
squared minus 18𝑡 plus 𝑐. This is known as the general
expression. We, however, know that its initial
velocity is 20 meters per second. So we can say that when 𝑡 is equal
to zero, 𝑣 is equal to 20.
And we can use this information to
find the particular expression for velocity. We substitute 𝑡 equals zero and 𝑣
equals 20 into this equation. And we obtain 20 equals zero squared
minus 18 times zero plus 𝑐, which gives us 20 equals 𝑐. And we have the expression for
velocity at time 𝑡. It’s 𝑡 squared minus 18𝑡 plus
20. We’re gonna integrate one more time
to find the expression for displacement. It’s the integral of 𝑡 squared
minus 18𝑡 plus 20 with respect to 𝑡.
This time, the integral of 𝑡
squared is 𝑡 cubed over three. The integral of negative 18𝑡 is
negative 18𝑡 squared over two. The integral of 20 is 20𝑡 and we
have a constant of integration. Notice I’ve called this 𝑑 instead
of 𝑐 because we’ve already used 𝑐 in this question. So 𝑠 is equal to 𝑡 cubed over
three minus nine 𝑡 squared plus 20𝑡 plus 𝑑. Once again, we have enough
information to work out the value of 𝑑. We know the initial displacement is
zero meters. So when 𝑡 is equal to zero, 𝑠 is
equal to zero. And we substitute these values
in. And we obtain zero equals zero
cubed over three minus nine times zero squared plus 20 times zero plus 𝑑, which
tells us that zero is equal to 𝑑. And we’re done! We found the expression for the
displacement of the particle at time 𝑡. It’s 𝑡 cubed over three minus nine
𝑡 squared plus 20𝑡 meters.
In our next example, we’re going to
look at how we can use integration to solve problems involving optimization.
A particle started moving in a
straight line. Its acceleration at time 𝑡 seconds
is given by 𝑎 equals negative five 𝑡 squared plus five meters per square seconds,
when 𝑡 is greater than or equal to zero. Find the maximum velocity of the
particle 𝑣 max and the distance 𝑥 it travelled before it attained this velocity,
given that the initial velocity of the particle is zero meters per second.
To answer this question, we begin
by recalling that we can find an expression for the velocity by integrating the
expression for acceleration with respect to time. Similarly, we can find an
expression for displacement by integrating the expression for velocity with respect
to time. We also need to recall that we can
find a maximum by first looking for the critical points of a function and these
occur at places where the derivative of that function is equal to zero or does not
exist.
Well, to find the critical points
and ultimately the maximum for the velocity, we want to work out when d𝑣 by d𝑡 is
equal to zero. But d𝑣 by d𝑡 is, of course,
acceleration. So let’s set our expression for 𝑎
equal to zero and solve for 𝑡. That’s negative five 𝑡 squared
plus five equals zero. Adding five 𝑡 squared to both
sides, we get five 𝑡 squared equals five. And then dividing through by five,
we obtain 𝑡 squared to be equal to one. Our final step is to take the
square root of both sides. And remember, we don’t need to
worry about finding the positive and negative square root of one since time must be
a positive value. So 𝑡 is equal to one. And we found that there’s a
critical point for our function for velocity, and it occurs when 𝑡 is equal to
one.
Since this is the only critical
point, it’s safe to assume it probably is a maximum. But we’ll check by finding the
second derivative. If the second derivative is less
than zero when 𝑡 is equal to one, then we will indeed have a maximum. And of course, the second
derivative of velocity with respect to time is equal to the first derivative of
acceleration with respect to time. Well, the first derivative of
negative five 𝑡 squared plus five is simply negative 10𝑡. We’re going to evaluate this when
𝑡 is equal to one. So that’s negative 10 times one,
which is, of course, negative 10. Since negative 10 is less than
zero, we do indeed have a maximum when 𝑡 is equal to one.
Now, we’re actually trying to find
the maximum velocity of the particle. And we know it occurs when 𝑡 is
equal to one. So let’s find the function of the
velocity and evaluate it when 𝑡 equals one. We said that this was the integral
of the function for acceleration. So we’re integrating negative five
𝑡 squared plus five. Well that gives us negative five 𝑡
cubed over three plus five 𝑡 plus its constant of integration 𝑐. We can work out what this constant
of integration is by using the fact that the initial velocity of the particle is
zero meters per second. In other words, when 𝑡 is equal to
zero, 𝑣 is equal to zero. Substituting these values in gives
us zero equals negative five times zero cubed over three plus five times zero plus
𝑐. And that tells us that zero is
equal to 𝑐. And our final expression for
velocity is negative five 𝑡 cubed over three plus five 𝑡.
We want to know the maximum
velocity. And we saw that this occurs when 𝑡
is equal to one. So we’re now going to substitute 𝑡
equals one into this expression. That’s negative five times one
cubed over three plus five times one, which is ten thirds. And we have the first part of our
solution. The max is equal to ten thirds. And it’s important to realize that
there’s no other maximum at the end point where 𝑡 is equal to zero, since there are
no other turning points in the interval zero to one.
We’re now going to work out the
distance 𝑥 it travelled before it attained this velocity. Now, we’ll need to be a little bit
careful here. Distance is the scalar version of
displacement. It’s the absolute value of
displacement. We know that we can integrate our
function for velocity to find a function for displacement. And then, we can evaluate that
between one and zero to find the total displacement. What we’ll do to check whether this
is going to give us the same value as distance is check the shape of the graph. If the curve sits purely above or
purely below the 𝑥-axis, then we know that the absolute value of the displacement
will be equal to the distance. Well, between zero and one, the
graph does indeed sit purely above the 𝑥-axis or the 𝑡-axis in this case.
So the total distance travelled
will be given by the area between the curve and the 𝑡-axis bounded by the lines 𝑡
equals zero and 𝑡 equals one. That’s simply the definite integral
between zero and one of the function for velocity with respect to time. Integrating gives us negative five
𝑡 to the fourth power over 12 plus five 𝑡 squared over two. And when we evaluate that between
one and zero, we get negative five over 12 plus five over two minus zero, which is
25 12ths. 𝑣 max is equal to ten thirds and
the distance 𝑥 it travelled before it attained 𝑣 max is 25 over 12 meters.
In this example, we saw that we can
use a definite integration to evaluate the total distance travelled of a
particle. We’ll consider one more example of
this form.
A particle moves along the 𝑥-axis
with a velocity 𝑣 meters per second of 𝑣 equals cos of 𝑡. Find the total distance travelled
during the time interval zero is less than or equal to 𝑡, which is less than or
equal to three 𝜋 by 𝑡.
We have to be careful here. Remember, distance is the magnitude
of displacement. We find displacement by evaluating
the integral of the function for velocity. Now that, of course, can be
considered as the area between the curve and the 𝑥-axis. The problem we have here is that
cos of 𝑡 is both positive and negative over time interval 𝑡 is greater than or
equal to zero and less than or equal to three 𝜋 by two. So we’re going to evaluate this in
two parts.
We’ll say that the total distance
travelled is equal to the sum of the magnitude of the displacement between 𝑡 equals
zero and 𝑡 equals 𝜋 by two and the magnitude of the displacement between 𝑡 equals
𝜋 by two and three 𝜋 by two. The magnitude of the displacement
between 𝑡 equals zero and 𝑡 equals 𝜋 by two is quite straightforward. It’s the definite integral of cos
of 𝑡 evaluated between zero and 𝜋 by two. The magnitude of the displacement
between 𝜋 by two and three 𝜋 by two is a little trickier though. Since this part of the graphs sits
below the 𝑥-axis, we know we’re going to end up with a negative value when we
integrate. We can, therefore, say that the
magnitude is equal to the negative integral of cos of 𝑡 between 𝜋 by two and three
𝜋 by two or the integral evaluated between three 𝜋 by two and 𝜋 by two of cos of
𝑡.
Remember, reversing our limits just
has the effect of changing the sign of our solution. The integral of cos of 𝑥 d𝑥 is
sin of 𝑥 plus 𝑐. So the integral of cos of 𝑡 is sin
of 𝑡. And we don’t need that constant of
integration because we’re dealing with a definite integral. Evaluating this between our limits
and we get sin of 𝜋 by two minus sin of zero plus sin of 𝜋 by two minus sin of
three 𝜋 by two. sin of 𝜋 by two is one and sin of zero is zero. We also know that sin of three 𝜋
by two is negative one. So we have one minus zero plus one
minus negative one, which is equal to three. And we can, therefore, say that the
total distance travelled during the time interval 𝑡 is greater than or equal to
zero and less than or equal to three 𝜋 by two is three meters.
In this video, we’ve seen that we
can use integration to derive functions for velocity and displacement from functions
for acceleration and velocity, respectively. We also saw that we can integrate a
function for acceleration twice to help us find a function for displacement, but
that we need initial values for velocity and displacement to find a particular
solution. We also saw that definite integrals
can be used to help us find the total displacement or distance travelled. But that’s in a latter case, we’ll
need to consider the shape of the graph before evaluating.
