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.
