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Studying for a test? Prepare with these 5 lessons on Applications of derivatives.

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# Related rates: Approaching cars

Video transcript

So this car right over here
is approaching an intersection at 60 miles per hour. And right now, right
at this moment, it is 0.8 miles from
the intersection. Now we have this
truck over here, it's approaching the
same intersection on a street that is
perpendicular to the street that the car is on. And right now it is 0.6
miles from the intersection. And it is approaching
the intersection at 30 miles per hour. Now my question to you
is, what is the rate at which the distance
between the car and the truck is changing? Well to think about
that, let's first just think about
what we're asking. So we're asking
about the distance between the car and the truck. So right at this
moment, when the car is 0.8 miles from
the intersection, the truck is 0.6 miles
from the intersection. The truck is traveling at
30 miles per hour towards intersection, the car is
travelling at 60 miles per hour towards the intersection
right at this moment. What is the rate at which
this distance right over here is changing? And just so that we have
some variables in place, let's call this distance s. So what we really are
trying to figure out is right at this moment, what
is ds dt going to be equal to? Let's think about
what we know that we could use to somehow
come to terms or figure out what ds dt is. Well we know the distance of
the car and the intersection. And let's just
call that distance, let's call that-- I don't know,
let's call that distance y. So y is equal to 0.8 miles. We also know that--
so let me write this-- we know that y
is 0.8 miles right now. We also know the dy
dt, the rate at which y is changing with
respect to time is what? Well y is decreasing
by 60 miles per hour. So let me write it as
negative 60 miles per hour. Now similarly, let's say that
this distance right over here is x. x is 0.6 miles
right at this moment. So we know that x is
equal to 0.6 miles. What is the rate at which x is
changing with respect to time? Well, we know it's
30 miles per hour is how fast we're
approaching the intersection, but x is decreasing by
30 miles every hour. So we should say it's
negative 30 miles per hour. So we know what y is. We know what x is. We know how fast y is
changing, how fast x is changing with
respect to time. So what we could try
to do here is come up with a relationship
between x, y, and s. And then differentiate
that relationship with respect to time. And it seems like we have
pretty much everything we need to solve for this. So what's a relationship
between x, y, and s? Well we know that this
is a right triangle. The streets are
perpendicular to each other. So we can use the
Pythagorean theorem. We know that x squared
plus y squared is going to be equal to s squared. And then we can
take the derivative of both sides of
this with respect to time to get a relationship
between all the things that we care about. So what's the derivative of x
squared with respect to time? Well, so you're going to need
the derivative of x squared with respect to x,
which is just 2x, times the derivative of x with
respect to time, times dx dt. Once again, just the chain rule. Derivative of something squared
with respect to the something, times the derivative of the
something with respect to time. And we use similar
logic right over here when we want to take
the derivative of y squared with respect to time. Derivative of y squared
with respect to y, times the derivative of
y with respect to time. Now on the right-hand
side of this equation, we once again take the
derivative with respect to time. So it's the derivative of s
squared with respect to s, which is just 2s. Times the derivative of
s with respect to time. Once again, this is all just an
application of the chain rule. So now it looks
like we know what x is, we know what
dx dt is, we know what y is, we know
what dy dt is. All we need to
figure out is what s and then what ds dt is, the
rate at which this distance is changing with respect to time. Well what's s right now? Well we can actually use
the Pythagorean theorem at this exact moment. We know that x
squared-- so x is 0.6-- we know that 0.6 squared
plus y squared, 0.8 squared is equal to s squared. Well this is 0.36, plus
0.64 is equal to s squared. This is 1, it is
equal to s squared. And we only care about
positive distances, so we have s is
equal to 1 right now. So we also know what s is. So let's substitute
all of these numbers in and then try to solve for
what we came here to do. Solve for ds dt. So the rate at which, so
2 times x-- maybe I'll do that in yellow--
2 times x, x is 0.6 so it's going to
be 1.2 times dx dt, so that's negative
30 miles per hour. So times negative
30 miles per hour, plus 2 times y is
1.6, times dy dt is negative 60 miles per hour. And I'm not writing
the units here. But if you were to
write the units, you will see that all of
our distances are in miles. And all of our time
is within hours, so we're going to
get an answer when we solve for ds dt
that's miles per hour. But I encourage you, if you
want to, to actually write out the units and see
how they work out. And so this is going to
be equal to 2 times s. Well, s is 1 mile, so it's
just going to be 2 times ds dt, which is what
we're trying to solve for. So what do we get here
on the left-hand side? So 1.2 times negative
30, that's negative 36. Right? 1/5 of 30 is 6,
yep that's right. And then 1.6 times
negative sixty, that's going to be negative
96, is equal to 2 times ds dt, is equal to 2 times
the rate at which our distance is changing
with respect to time. On the left-hand side right
over here, this is negative 132. Negative 132 is equal
to 2 times ds dt. Divide both sides by
2, we get negative 66. And now we could put our units
if we want, miles per hour is the rate at
which our distance is changing with
respect to time. So ds dt is negative
66 miles per hour. Does it make sense that we
got a negative number here? Well sure, this distance
is decreasing right at this moment, as they
approach the intersection.