### Video Transcript

In this video, weβll learn how to
describe motion of a particle along a curve defined by parametric functions. Weβll consider these in terms of
displacement, velocity, and acceleration, as well as considering the magnitude of
each of these, and look how calculus with parametric equations can help us to solve
these problems. We recall that parametric equations
allow us to express π₯ and π¦ in terms of a third parameter. We usually use π‘ as it represents
time. Then we obtain the parametric
equations π₯ is equal to π of π‘ and π¦ is equal to π of π‘. Weβll also need to recall some
rules for motion. The first is if π is a function
for displacement at time π‘, π of π‘, then the rate of change of displacement with
respect to time, π prime of π‘, gives us the velocity at time π‘. Similarly, the rate of change of
velocity with respect to time, π£ prime of π‘, gives us the acceleration at time
π‘. Thatβs π of π‘. Armed with this information, letβs
have a look at a variety of problems that involve motion in a plane.

A particle has a position defined
by the equations π₯ equals π‘ cubed minus five π‘ and π¦ equals three minus two π‘
squared. Find the velocity vector of the
particle at π‘ equals two.

Weβre told that the position of the
particle is defined by a pair of parametric equations. In real terms, this means that,
given a value of time π‘, we obtain a coordinate pair π₯π¦ for the position of the
particle at that time. We could say that, in vector terms,
the position of the particle at time π‘ is π of π‘ equals π‘ cubed minus five π‘ π
plus three minus two π‘ squared π. This isnβt though quite what weβre
looking for. We want to find the velocity vector
of our particle when π‘ is equal to two. So we recall that, given a function
for displacement which of course is essentially the difference in the objectβs
position from one time to another, we find a function for velocity by
differentiating with respect to π‘.

This means that we can achieve a
velocity vector for our particle at time π‘ by differentiating each component for
the position vector with respect to π‘. Thatβs the derivative of π‘ cubed
minus five π‘. And then we differentiate the
vertical component as well, the derivative of three minus two π‘ squared. We know that to differentiate a
polynomial term, we multiply the entire term by the experiment and then reduce the
exposure by one. So π‘ cubed differentiates to three
π‘ squared and negative five π‘ differentiates to negative five.

Similarly, the derivative of three
minus two π‘ squared is negative four π‘. Remember, the derivative of any
constant is simply zero. Our vector function for velocity is
three π‘ squared minus five π plus negative four π‘ π. Remember, we want a velocity vector
at π‘ is equal to two. So weβre going to substitute π‘
equals two into our vector. Thatβs three times two squared
minus five π plus negative four times two π, which simplifies to seven π minus
eight π.

In our next example, weβll look at
how a subtle change in language can dramatically change our solution.

A particle has a position defined
by the equations π₯ equals five π‘ squared plus four π‘ and π¦ equals three π‘ minus
two. Find the speed of the particle at
π‘ equals two.

Here weβve been given a pair of
parametric equations to describe the position of the particle. This means that, given a value of
π‘, we obtain an output which is a coordinate pair for the position of the particle
at that time. We could treat these as completely
separate entities. Now, weβre looking to find the
speed of the particle at π‘ equals two. Well, we know that speed is the
magnitude of the velocity. And we also recall that we can find
a function for velocity by differentiating a function for displacement. We might then choose to say that,
in vector terms, the position of the particle at time π‘ is five π‘ squared plus
four π‘ π plus three π‘ minus two π. We could then differentiate our
function for position and find a function for velocity. And in fact, we can achieve this by
differentiating individually the horizontal and vertical components for our
displacement or our position.

Weβll begin by differentiating the
horizontal component. Thatβs five π‘ squared plus four
π‘. The derivative of five π‘ squared
is 10π‘. And the derivative of four π‘ with
respected π‘ is simply four. Letβs repeat this process for the
vertical component. The derivative of three π‘ is three
and the derivative of negative two is zero. In vector terms then, our velocity
is described by 10π‘ plus four π plus three π. All this means is, at π‘ seconds,
the velocity is 10π‘ plus four in the horizontal direction and three in the
vertical, the π direction. Next, we want to work out the
velocity at π‘ equals two. So weβll substitute π‘ into our
vector equation for velocity that gives us 10 times two plus four π plus three
π. Thatβs 24π plus three π.

Remember that weβre looking to find
the speed, which is the magnitude of the velocity. So we recall that we can find the
magnitude of a vector π given as π₯ π plus π¦ π by finding the square root of π₯
squared plus π¦ squared. In this case, thatβs the square
root of 24 squared plus three squared, which is equal to three root 65. Now, we donβt have any units here,
so weβre done. The speed of the particle at π‘
equals two is three root 65.

Weβll now consider how a similar
process can help us to evaluate acceleration.

A moving particle is defined by the
two equations π₯ equals π‘ cubed minus five π‘ minus five and π¦ equals seven π‘
squared minus three. Find the magnitude of the
acceleration of the particle at π‘ equals one.

In this question, weβve been given
the position of the particle as defined by a pair of parametric equations. So given a value of π‘, we obtain a
coordinate pair π₯π¦ for the position of our particle. We could choose to consider this in
vector terms and say that the position of the particle at time π‘, π of π‘, is
given by π‘ cubed minus five π‘ minus five π plus seven π‘ squared minus three
π. Now, in this question, weβre
looking to find the acceleration. Well, in fact, we want the
magnitude of the acceleration, but weβll deal with that in a moment. So we recall that the acceleration
is equal to the derivative of the velocity with respect to time. But we also know that the velocity
is equal to the first derivative of the displacement or the position vector.

We can in turn say that, to find a
function for acceleration, weβre going to need to differentiate our function for
position twice with respect to time. Weβll do it once to find the
function for velocity. We can differentiate each component
function in turn. When we differentiate π‘ cubed
minus five π‘ minus five, we get three π‘ squared minus five. And when we differentiate seven π‘
squared minus three, we get 14π‘. So our function for velocity is
three π‘ squared minus five π plus 14π‘ π. But what does this actually
mean? Well, it means that the velocity
can be defined in terms of its horizontal velocity and its vertical velocity. Horizontally, its velocity is three
π‘ squared minus five, but vertically, itβs given by the function 14π‘.

Okay, great. Letβs differentiate again to find
our function for acceleration. And when we differentiate three π‘
squared minus five, we find that the acceleration in the horizontal direction is six
π‘. We then differentiate 14π‘. And we see that the acceleration in
the vertical direction is 14. Weβre now able to find a vector
acceleration for our particle at π‘ equals one. We simply substitute π‘ equals one
into a vector function for acceleration. And we find that the acceleration
at π‘ equals one is given by the vector six π plus 14π. We, of course, want to find the
magnitude of the acceleration though.

So we recall that the magnitude of
a vector in two dimensions given by π₯ π plus π¦π is the square root of π₯ squared
plus π¦ squared. This means the magnitude of our
acceleration at π‘ equals one is the square root of six squared plus 14 squared,
which is two root 58. There are no units here. So we found the magnitude of the
acceleration of the particle at π‘ equals one to be two root 58.

We can also use this process to
solve more complicated problems involving planar motion. Letβs see what that might look
like.

If a particle is moving on a curve
defined by the parametric equations. π₯ is equal to a half π‘ squared
minus four π‘ plus three and π¦ is equal to a half π‘ squared plus three π‘. Find the time to the nearest tenth,
at which π£ equals 64.

Weβre given a pair of parametric
equations which describe the position or displacement of our particle. And weβre looking to find the time
at which the velocity is equal to 64. We could think about this in vector
terms. And we could say that the
displacement is equal to a half π‘ squared minus four π‘ plus three in the
horizontal direction, so thatβs π, and a half π‘ squared plus three π‘ in the
vertical direction. So that part is π. We also know that we can
differentiate the function for displacement and weβll achieve a function for
time. Well, here we can achieve that by
differentiating each of our component functions of the vector for displacement. When we differentiate a half π‘
squared minus four π‘ plus three, we end it with simply π‘ minus four. And when we differentiate a half π‘
squared plus three π‘, we had π‘ plus three.

So we now know that the velocity of
the particle in the horizontal direction at time π‘ is π‘ minus four. And in the vertical direction, itβs
π‘ plus three. And weβve represented that in
vector form. We want to know the time at which
π£ is equal to 64. Notice that this is not given as a
vector quantity. So weβre going to need to work out
when the magnitude of our velocity is equal to 64. Weβll use the fact that the
magnitude of a two-dimensional vector π₯π plus π¦π is equal to the square root of
π₯ squared plus π¦ squared. And so this means the magnitude of
our velocity is the square root of π‘ minus four all squared plus π‘ plus three all
squared. Now, of course, this is equal to
64. All we need to do now is solve for
π‘.

We begin by squaring both sides of
our equation and we find π‘ minus four all squared plus π‘ plus three all squared
equals 4096. We distribute each pair of
parentheses. And we see we obtain π‘ squared
minus eight π‘ plus 16 plus π‘ squared plus six π‘ plus nine on the left-hand
side. And that simplifies really nicely
to two π‘ squared minus two π‘ plus 25 equals 4096. We achieve a quadratic equation
that we can solve by subtracting 4096 from both sides. Itβs two π‘ squared minus two π‘
minus 4071 equals zero. And using any method available to
us, we can solve this quadratic equation. We might use completing the square
or the quadratic formula. And we might even use a polynomial
equation solver on a calculator. When we do, we achieved π‘ equals
45.6192 or negative 44.6192. So which do we choose? Well, this is time, so it makes no
sense to have this negative value. And so corrected to the nearest
tenth, the time at which π£ equals 64 is 45.6.

In our final example, weβll
consider an initial value problem.

Suppose that a particle is moving
on a curve defined by the parametric equations dπ₯ by dπ‘ equals five π‘ minus 15
and dπ¦ by dπ‘ equals eight minus four π‘. If the particle is initially at
horizontal displacement π equals 32.3, find the minimum horizontal displacement
from π equals zero.

The motion of our particle is
described by a pair of parametric differential equations. We have dπ₯ by dπ‘ equals five π‘
minus 15 and dπ¦ by dπ‘ equals eight minus four π‘. Letβs think about what those
equations are actually representing. Well, if π₯ and π¦ are functions
that describe the position of the particle in terms of horizontal and vertical
components, then dπ₯ by dπ‘ and dπ¦ by dπ‘ must represent velocity functions, again,
in both the horizontal and vertical direction. Weβre looking to find the
displacement of our particle. But thatβs not just it. We need to find the minimum
horizontal displacement. So we recall two pieces of
information.

Firstly, we know that if we
differentiate a function for displacement, we achieve a function for velocity. Conversely, we can say that if we
integrate a function for velocity with respect to time, weβll achieve a function for
displacement. In this case, weβll achieve a
function that describes the horizontal displacement by integrating our function for
velocity in the horizontal direction. Itβs the integral of dπ₯ by dπ‘
with respect to π‘. Well, thatβs the integral of five
π‘ minus 15. We also know that when we integrate
a polynomial term whose exponent is not equal to negative one, we add one to the
exponent and then divide by that new value. So the integral of five π‘ is five
π‘ squared over two, and the integral of negative 15 is negative 15π‘. Weβre dealing with an indefinite
integral. So we add a constant of
integration, which Iβve called π.

Weβre also told, though, that the
particle is initially at a horizontal displacement of π equals 32.3, well,
initially means when π‘ is equal to zero. And so we can substitute π‘ equals
zero and π equals 32.3 into our equation for π π₯ of π‘. And we find that 32.3 equals five
times zero squared over two minus 15 times zero plus π. Well, five times zero squared over
two and negative 15 times zero are both equal to zero. So π is equal to 32.3. And we have an expression for the
horizontal displacement of our particle. Itβs five π‘ squared over two minus
15π‘ plus 32.3. Weβre looking to find the minimum
horizontal displacement. And so we recall that we can find
the critical point of a function by differentiating that function and setting it
equal to zero.

In our case, weβre interested in
the minimum horizontal displacement. The derivative of the displacement
in the horizontal direction is dπ₯ by dπ‘. And thatβs five π‘ minus 15. So weβll set this equal to zero to
find the location of any critical points. We solve by adding 15 to both sides
and dividing through by five. And we find that π‘ equals three is
a critical point of our horizontal function. To establish whether itβs a
minimum, we need to work out whether the second derivative at the point where π‘
equals three is greater than zero.

So weβre going to differentiate our
function dπ₯ by dπ‘. When we do, we see that five π‘
minus 15 differentiates to five. Five is greater than zero. That tells us that all critical
points that might occur must be a local minimum. So, specifically, when π‘ equals
three, we have a local minimum. In fact, this is a global minimum,
since our equation for π π₯ of π‘ is a quadratic with a positive leading
coefficient. And this means thereβs just one
critical point on our curve. To work out then the minimum
horizontal displacement from π equals zero, weβll substitute π‘ equals three into
our expression for the horizontal displacement. Thatβs five times three squared
over two minus 15 times three plus 32.3, which equals 9.8. The minimum horizontal displacement
from π equals zero is 9.8.

Noticing this question, we never
actually considered the equation that describes the vertical motion of our
particle. And thatβs because we were only
interested in the horizontal displacement. In this video, we saw that by
considering parametric equations is a vector function or simply considering their
horizontal and vertical components, we can deal effectively with planar motion. We also saw that we can use the
standard rules weβre so used to for calculus and motion to help us convert between
displacement, velocity, and acceleration given in parametric equation form.