Video Transcript
A particle is moving along the
curve six 𝑦 squared plus two 𝑥 squared minus two 𝑥 plus five 𝑦 minus 13
equals zero. If the rate of change of its
𝑥-coordinate with respect to time as it passes through the point negative one,
three is two, find the rate of change of its 𝑦-coordinate with respect to time
at the same point.
So let’s begin by looking at
what the question actually wants us to find. It wants us to find the rate of
change of its 𝑦-coordinate. Now we recall that the rate of
change of a quantity is its derivative with respect to time. So in this case, we’re looking
to find d𝑦 by d𝑡. And in fact, we’re given
information about the rate of change of the 𝑥-coordinate with respect to
time. We’re given information about
d𝑥 by d𝑡. At a given point, at the point
negative one, three, the value of d𝑥 by d𝑡 is equal to two. And so what we’re going to do
first is begin by differentiating the entire equation of our curve. We’ll need to use implicit
differentiation to do so.
Remember, we’re differentiating
with respect to 𝑡. And the derivative of zero is,
of course, zero. We can then go through and
differentiate term by term. And of course, we’re going to
use implicit differentiation to do so. Let’s begin with six 𝑦
squared. We can differentiate six 𝑦
squared with respect to 𝑦. It will be two times six
𝑦. That simplifies to 12𝑦. And so that means the
derivative of this expression with respect to 𝑡 will be 12𝑦 times d𝑦 by
d𝑡. We perform a similar process
for two 𝑥 squared. We differentiate it with
respect to 𝑥. And that’s two times two 𝑥 or
four 𝑥. And then we multiply that by
d𝑥 by d𝑡.
Next, we differentiate negative
two 𝑥 with respect to 𝑡. And to do so, we differentiate
with respect to 𝑥 and then times that by d𝑥 by d𝑡. In a similar way, the
derivative of five 𝑦 with respect to 𝑡 is five d𝑦 by d𝑡. Then the derivative of our
constant, negative 13, is simply equal to zero. We’re now going to factor for
d𝑦 by d𝑡 and d𝑥 by d𝑡 just to neat some things up a little bit. When we do, we find that d𝑦 by
d𝑡 times 12𝑦 plus five plus d𝑥 by d𝑡 times four 𝑥 minus two is equal to
zero.
Since we’re trying to find the
value of d𝑦 by d𝑡 at a given point, we’re going to rearrange to make d𝑦 by
d𝑡 the subject. We begin by subtracting the
term containing d𝑥 by d𝑡 from both sides; then we divide through by 12𝑦 plus
five. And so our expression for d𝑦
by d𝑡 is as shown. All that’s left is to perform a
few substitutions. Firstly, we know the value of
d𝑥 by d𝑡 is equal to two. And we’re interested in finding
the rate of change of the 𝑦-coordinate with respect to time at the same point,
when 𝑥 is equal to negative one and 𝑦 is equal to three.
And so we substitute all of
these values into our expression for d𝑦 by d𝑡. We get negative two times four
times negative one minus two over 12 times three plus five. And that gives us a value of 12
over 41. We aren’t given any units for
the rate of change of the 𝑥-coordinate. Although, we might assume that
it’s units per time. And so the rate of change of
the 𝑦-coordinate with respect to time at the same point is 12 over 41 or 12
over 41 units per time.
