Application: Related Rates
We use the chain rule and implicit differentiation to pry into rates that are related to each other.
Usually in a real-world (or even non-real-world!) situation, several quantities are changing in a
complicated connection with each other. Sometimes we know how one quantity is changing and
we want to know how a related quantity changes. The goal is to set up equations relating the
various quantities, differentiate everything in sight (remembering the chain rule and implicit
differentiation!) and solve for the thing you need. Examples are the best place to start.
Example: when an object falls, it gains speed. When it then hits the ground, the kinetic energy
is changed into other forms of energy (like sound, etc.). The energy released is mv2/2, where m
is the mass of the object and v is the speed when it hits the ground.
Let’s say you drop a 5kg mass so that it hits the ground at a speed of 20m/s, so the energy
released is 5⋅202/2 = 1000 J. If we let it drop a little farther, at what rate is the energy released
increasing with extra time dropping?
Solution: here, we can actually figure things out directly. We know that v = at where a is the
acceleration, 9.8m/s2 due to gravity. So we plug in and get E = mv2/2 = m(at)2/2 = ma2t2/2.
Differentiating gives dE/dt = ma2t. Now we know m = 5kg, and at = the speed from before, so is
20m/s. So we get dE/dt = 5⋅a⋅20 = 980 J/s. In other words, at this particular speed, for each
additional second of drop time, we’d get another 980 joules of energy out.
Here, we could solve for the quantity we wanted (energy) directly in terms of the quantity we
could control (time), so we used a direct formula for E in terms of t to find dE/dt. You can’t
always solve for the quantity you want, though, so we you have be a little more creative.
Helicopter
Example: (from OZ, 2nd ed.) a police helicopter
hovers 1000 feet above a highway where the
speed limit is 65mph. A car is spotted on the
freeway below, and when the office activates the
2000
radar gun a moment later the radar pulse indicates 1000
the car is 2000 feet away, receding at 85 feet per
second. Is the car speeding?
Solution: normally, you’d think that 60mph = 88
feet per second, so 85fps can’t be speeding. But
the car is not moving directly away from the helicopter—the angle might make a difference.
Let’s look deeper. Let x = horizontal distance car has traveled along the road, y = 1000ft =
vertical distance of helicopter from road, and z = distance car is from helicopter. Currently this
is 2000ft, but it is clearly changing. In fact, we know dz/dt = 85fps. We want to know the speed
of the car along the road, which is dx/dt. And what is the relationship between these quantities?
Pythagoras to the rescue! We know x2 + y2 = z2, or x2 + 10002 = z2. Note that even though we
know the current value of z, we don’t put it in, because z is changing. Now we could subtract
10002 from both sides, take the square root, and thus solve for x. But that is kind of messy.
Let’s just differentiate: 2x dx/dt + 0 = 2z dz/dt. So the quantity we want, dx/dt = (z/x) dz/dt. We
can no substitute known quantities: z = 2000 feet, dz/dt = 85fps, and x can be found from the
original Pythagorean equation (if we put in z-right-now = 2000 feet, we find x-right-now =
1000 3 feet). So dx/dt = 2000⋅85 / 1000 3 fps, which works out to about 98.15fps. So the car
is traveling much faster along the road than it is just away from the helicopter! In fact, 98.15fps
is about 66.92mph, so the car is speeding, though just by a tiny bit!
750
Example: A lighthouse has a beam that revolves once every
Lighthouse
20 seconds. The lighthouse is on an island 750 meters from
the short. If you were 1000 meters down shore from the point
straight across from the lighthouse, how fast would you have to
1000
travel to keep up with the light?
Solution: Let’s let θ be the angle that the light has rotated
through. We know that θ is changing at a rate of 1 revolution
per 20 seconds, or (1 rev / 20 sec)(2π radians / revolution) =
π/10 radians/second. We also know that the distance along the
shore that the light has traveled is changing, though the 750 meters from the shore to the
lighthouse isn’t. Let the distance along the shore be called x; currently x = 1000 meters. But
from simple trig, x / 750 = tan(θ), or x = 750 tan(θ). We could substitute for θ now, or just
differentiate: dx/dt = 750sec2(θ)⋅dθ/dt. Now the current value of θ will give a value of 25/9 for
sec2(θ) (the hypotenuse in the triangle above is 1250 meters). So dx/dt = 750⋅25/9⋅π/10, which is
625π/3 meters/second, or about 654.5m/s, or about 1464 miles per hour. Run fast!
Example: an equation from optics says that for a certain type of telescope 1/i – 1/o = 1/f. In this
equation, i is the image distance, or how far away the image appears. The value of o is the
object distance, how far away the object actually is. And f is the focal distance, which is a fixed
property of the telescope and its lenses. The object in view is 500 meters away, but appears only
20 meters away in our telescope. If the object is approaching us at a rate of 2 meters/second,
how quickly does the image appear to approach us?
Solution: differentiate: – (di/dt)/i2 + (do/dt)/o2 = 0. Thus, di/dt = i2/o2 do/dt. Using the values of
i = 20, o = 500, and do/dt = –2 (it is getting closer!), we get di/dt = –8/2500 m/s, so the image
appears to be getting closer at a fairly slow rate.
The steps to solving these kinds of problems are:
1) If possible, draw a picture!
2) Write equations that involve all the known quantities in the problem. Do not plug in values
for these quantities unless they do not change.
3) Differentiate everything in sight, remembering the chain rule and implicit differentiation.
4) Solve for the derivative you need.
5) Plug in current value for the rest of the variables. You may need to plug them in to your
original, un-differentiated equations, too, to solve for some of the quantities which can be
determined but whose values aren’t listed (like the distance along the road in the speeding
example).
Try the practice problems!