# Question Video: Finding the Area of a Region Bounded by Two Quadratic Functions Mathematics • Higher Education

Suppose that a particle is moving on a curve defined by the parametric equations d𝑥/d𝑡 = 5𝑡 − 15 and d𝑦/d𝑡 = 8 − 4𝑡. If the particle is initially at horizontal displacement 𝑑 = 32.3, find the minimum horizontal displacement from 𝑑 = 0.

03:50

### Video Transcript

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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.