# Video: Planar Motion Using Parametric Equations

In this video, we will learn how to describe motion of a particle along a curve defined by parametric functions.

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### Video Transcript

In this video, we’ll learn how to describe motion of a particle along a curve defined by parametric functions. We’ll 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 𝑥 and 𝑦 in terms of a third parameter. We usually use 𝑡 as it represents time. Then we obtain the parametric equations 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡. We’ll also need to recall some rules for motion. The first is if 𝑠 is a function for displacement at time 𝑡, 𝑠 of 𝑡, then the rate of change of displacement with respect to time, 𝑠 prime of 𝑡, gives us the velocity at time 𝑡. Similarly, the rate of change of velocity with respect to time, 𝑣 prime of 𝑡, gives us the acceleration at time 𝑡. That’s 𝑎 of 𝑡. Armed with this information, let’s have a look at a variety of problems that involve motion in a plane.

A particle has a position defined by the equations 𝑥 equals 𝑡 cubed minus five 𝑡 and 𝑦 equals three minus two 𝑡 squared. Find the velocity vector of the particle at 𝑡 equals two.

We’re 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 𝑡, we obtain a coordinate pair 𝑥𝑦 for the position of the particle at that time. We could say that, in vector terms, the position of the particle at time 𝑡 is 𝑠 of 𝑡 equals 𝑡 cubed minus five 𝑡 𝑖 plus three minus two 𝑡 squared 𝑗. This isn’t though quite what we’re looking for. We want to find the velocity vector of our particle when 𝑡 is equal to two. So we recall that, given a function for displacement which of course is essentially the difference in the object’s position from one time to another, we find a function for velocity by differentiating with respect to 𝑡.

This means that we can achieve a velocity vector for our particle at time 𝑡 by differentiating each component for the position vector with respect to 𝑡. That’s the derivative of 𝑡 cubed minus five 𝑡. And then we differentiate the vertical component as well, the derivative of three minus two 𝑡 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 𝑡 cubed differentiates to three 𝑡 squared and negative five 𝑡 differentiates to negative five.

Similarly, the derivative of three minus two 𝑡 squared is negative four 𝑡. Remember, the derivative of any constant is simply zero. Our vector function for velocity is three 𝑡 squared minus five 𝑖 plus negative four 𝑡 𝑗. Remember, we want a velocity vector at 𝑡 is equal to two. So we’re going to substitute 𝑡 equals two into our vector. That’s three times two squared minus five 𝑖 plus negative four times two 𝑗, which simplifies to seven 𝑖 minus eight 𝑗.

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

A particle has a position defined by the equations 𝑥 equals five 𝑡 squared plus four 𝑡 and 𝑦 equals three 𝑡 minus two. Find the speed of the particle at 𝑡 equals two.

Here we’ve been given a pair of parametric equations to describe the position of the particle. This means that, given a value of 𝑡, 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, we’re looking to find the speed of the particle at 𝑡 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 𝑡 is five 𝑡 squared plus four 𝑡 𝑖 plus three 𝑡 minus two 𝑗. 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.

We’ll begin by differentiating the horizontal component. That’s five 𝑡 squared plus four 𝑡. The derivative of five 𝑡 squared is 10𝑡. And the derivative of four 𝑡 with respected 𝑡 is simply four. Let’s repeat this process for the vertical component. The derivative of three 𝑡 is three and the derivative of negative two is zero. In vector terms then, our velocity is described by 10𝑡 plus four 𝑖 plus three 𝑗. All this means is, at 𝑡 seconds, the velocity is 10𝑡 plus four in the horizontal direction and three in the vertical, the 𝑗 direction. Next, we want to work out the velocity at 𝑡 equals two. So we’ll substitute 𝑡 into our vector equation for velocity that gives us 10 times two plus four 𝑖 plus three 𝑗. That’s 24𝑖 plus three 𝑗.

Remember that we’re looking to find the speed, which is the magnitude of the velocity. So we recall that we can find the magnitude of a vector 𝑎 given as 𝑥 𝑖 plus 𝑦 𝑗 by finding the square root of 𝑥 squared plus 𝑦 squared. In this case, that’s the square root of 24 squared plus three squared, which is equal to three root 65. Now, we don’t have any units here, so we’re done. The speed of the particle at 𝑡 equals two is three root 65.

We’ll now consider how a similar process can help us to evaluate acceleration.

A moving particle is defined by the two equations 𝑥 equals 𝑡 cubed minus five 𝑡 minus five and 𝑦 equals seven 𝑡 squared minus three. Find the magnitude of the acceleration of the particle at 𝑡 equals one.

In this question, we’ve been given the position of the particle as defined by a pair of parametric equations. So given a value of 𝑡, we obtain a coordinate pair 𝑥𝑦 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 𝑡, 𝑠 of 𝑡, is given by 𝑡 cubed minus five 𝑡 minus five 𝑖 plus seven 𝑡 squared minus three 𝑗. Now, in this question, we’re looking to find the acceleration. Well, in fact, we want the magnitude of the acceleration, but we’ll 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, we’re going to need to differentiate our function for position twice with respect to time. We’ll do it once to find the function for velocity. We can differentiate each component function in turn. When we differentiate 𝑡 cubed minus five 𝑡 minus five, we get three 𝑡 squared minus five. And when we differentiate seven 𝑡 squared minus three, we get 14𝑡. So our function for velocity is three 𝑡 squared minus five 𝑖 plus 14𝑡 𝑗. 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 𝑡 squared minus five, but vertically, it’s given by the function 14𝑡.

Okay, great. Let’s differentiate again to find our function for acceleration. And when we differentiate three 𝑡 squared minus five, we find that the acceleration in the horizontal direction is six 𝑡. We then differentiate 14𝑡. And we see that the acceleration in the vertical direction is 14. We’re now able to find a vector acceleration for our particle at 𝑡 equals one. We simply substitute 𝑡 equals one into a vector function for acceleration. And we find that the acceleration at 𝑡 equals one is given by the vector six 𝑖 plus 14𝑗. 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 𝑥 𝑖 plus 𝑦𝑗 is the square root of 𝑥 squared plus 𝑦 squared. This means the magnitude of our acceleration at 𝑡 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 𝑡 equals one to be two root 58.

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

If a particle is moving on a curve defined by the parametric equations. 𝑥 is equal to a half 𝑡 squared minus four 𝑡 plus three and 𝑦 is equal to a half 𝑡 squared plus three 𝑡. Find the time to the nearest tenth, at which 𝑣 equals 64.

We’re given a pair of parametric equations which describe the position or displacement of our particle. And we’re 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 𝑡 squared minus four 𝑡 plus three in the horizontal direction, so that’s 𝑖, and a half 𝑡 squared plus three 𝑡 in the vertical direction. So that part is 𝑗. We also know that we can differentiate the function for displacement and we’ll 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 𝑡 squared minus four 𝑡 plus three, we end it with simply 𝑡 minus four. And when we differentiate a half 𝑡 squared plus three 𝑡, we had 𝑡 plus three.

So we now know that the velocity of the particle in the horizontal direction at time 𝑡 is 𝑡 minus four. And in the vertical direction, it’s 𝑡 plus three. And we’ve represented that in vector form. We want to know the time at which 𝑣 is equal to 64. Notice that this is not given as a vector quantity. So we’re going to need to work out when the magnitude of our velocity is equal to 64. We’ll use the fact that the magnitude of a two-dimensional vector 𝑥𝑖 plus 𝑦𝑗 is equal to the square root of 𝑥 squared plus 𝑦 squared. And so this means the magnitude of our velocity is the square root of 𝑡 minus four all squared plus 𝑡 plus three all squared. Now, of course, this is equal to 64. All we need to do now is solve for 𝑡.

We begin by squaring both sides of our equation and we find 𝑡 minus four all squared plus 𝑡 plus three all squared equals 4096. We distribute each pair of parentheses. And we see we obtain 𝑡 squared minus eight 𝑡 plus 16 plus 𝑡 squared plus six 𝑡 plus nine on the left-hand side. And that simplifies really nicely to two 𝑡 squared minus two 𝑡 plus 25 equals 4096. We achieve a quadratic equation that we can solve by subtracting 4096 from both sides. It’s two 𝑡 squared minus two 𝑡 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 𝑡 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 𝑣 equals 64 is 45.6.

In our final example, we’ll consider an initial value problem.

Suppose that a particle is moving on a curve defined by the parametric equations d𝑥 by d𝑡 equals five 𝑡 minus 15 and d𝑦 by d𝑡 equals eight minus four 𝑡. If the particle is initially at horizontal displacement 𝑑 equals 32.3, find the minimum horizontal displacement from 𝑑 equals zero.

The motion of our particle is described by a pair of parametric differential equations. We have d𝑥 by d𝑡 equals five 𝑡 minus 15 and d𝑦 by d𝑡 equals eight minus four 𝑡. Let’s think about what those equations are actually representing. Well, if 𝑥 and 𝑦 are functions that describe the position of the particle in terms of horizontal and vertical components, then d𝑥 by d𝑡 and d𝑦 by d𝑡 must represent velocity functions, again, in both the horizontal and vertical direction. We’re looking to find the displacement of our particle. But that’s 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, we’ll achieve a function for displacement. In this case, we’ll achieve a function that describes the horizontal displacement by integrating our function for velocity in the horizontal direction. It’s the integral of d𝑥 by d𝑡 with respect to 𝑡. Well, that’s the integral of five 𝑡 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 𝑡 is five 𝑡 squared over two, and the integral of negative 15 is negative 15𝑡. We’re dealing with an indefinite integral. So we add a constant of integration, which I’ve called 𝑎.

We’re also told, though, that the particle is initially at a horizontal displacement of 𝑑 equals 32.3, well, initially means when 𝑡 is equal to zero. And so we can substitute 𝑡 equals zero and 𝑠 equals 32.3 into our equation for 𝑠 𝑥 of 𝑡. And we find that 32.3 equals five times zero squared over two minus 15 times zero plus 𝑎. Well, five times zero squared over two and negative 15 times zero are both equal to zero. So 𝑎 is equal to 32.3. And we have an expression for the horizontal displacement of our particle. It’s five 𝑡 squared over two minus 15𝑡 plus 32.3. We’re 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, we’re interested in the minimum horizontal displacement. The derivative of the displacement in the horizontal direction is d𝑥 by d𝑡. And that’s five 𝑡 minus 15. So we’ll 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 𝑡 equals three is a critical point of our horizontal function. To establish whether it’s a minimum, we need to work out whether the second derivative at the point where 𝑡 equals three is greater than zero.

So we’re going to differentiate our function d𝑥 by d𝑡. When we do, we see that five 𝑡 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 𝑡 equals three, we have a local minimum. In fact, this is a global minimum, since our equation for 𝑠 𝑥 of 𝑡 is a quadratic with a positive leading coefficient. And this means there’s just one critical point on our curve. To work out then the minimum horizontal displacement from 𝑑 equals zero, we’ll substitute 𝑡 equals three into our expression for the horizontal displacement. That’s five times three squared over two minus 15 times three plus 32.3, which equals 9.8. The minimum horizontal displacement from 𝑑 equals zero is 9.8.

Noticing this question, we never actually considered the equation that describes the vertical motion of our particle. And that’s 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 we’re so used to for calculus and motion to help us convert between displacement, velocity, and acceleration given in parametric equation form.