Calculus > Related Rates

Related rate problems involve equations where a relationship exists between **two or more derivatives**. For example, you might want to find out the rate that the distance is increasing between two airplanes. Solving related rate problems has many real life applications. For example, a gas tank company might want to know the rate at which a tank is filling up, or an environmentalist might be concerned with the rate at which a certain marshland is flooding. Solving the problems usually involves knowledge of geometry and algebra in addition to calculus. Often, the “hard” part is the geometry or algebra — not the calculus, so you’ll want to make sure you brush up on those skills. Solving related relate problems also involves applications of the chain rule and implicit differentiation — where you differentiate both sides of the equation.

## Related Rates Sample problem #1

A rock is dropped into the center of a circular pond. The ripple moved outward at 4 m/s. How fast does the area change, with respect to time, when the ripple is 3m from the center?

Step 1: **Draw a picture of the problem **(this *always* helps, especially when geometry is involved).

Step 2: **Write out what you know about the problem**, using equations. You know that the rate at which the ripple (r, the radius of the circle) is moving, with respect to time *t*, is 4 m/s, so:

dr/dt = 4 m/s

Step 3: **Write out what you want to know **(what you are trying to solve for). You want to know how fast the area

*A*is changing with respect to time

*t*:

dA/dt = x, when r=3m

Step 4:

**Use the chain rule to find a solution for your Step 3 equation.**The chain rule tells us that:

dA/dt = dA/dr * dr/dt

Step 5:

**Figure out what dA/dr is.**From geometry, we know that A=πr

^{2}. So dA/dr is just the derivative of A=πr

^{2}.

d/dr[A] = d/dr[πr

^{2}]

= dA/dr = 2πr

So:

dA/dt = 2πr * dr/dt

Step 6:

**Solve the Step 5 equation.**You know that dr/dt (from Step 2) is 4 m/s, and r is 3 m (from the question), so:

dA/dt = 2π(3) * 4 m/s

= 24π m

^{2}/s

*That’s it!*

## Related Rates Sample problem #2:

The length of a rectangular drainage pond is changing at a rate of 8 ft/hr and the perimeter of the pond is changing at a rate of 24 ft/hr. At what rate is the width changing?

Step 1: **Figure out which geometric formulas are related **to the problem. From basic geometry, the formula for perimeter is P = (2*l) + (2*w) and that A = l*w.

Step 2: **Differentiate the perimeter equation**:

dp/dt = (2*dl/dt) + (2*dw/dt)

Step 3: Substitute in the information you know from the question. You know that the rate at which the perimeter is changing is 24 ft/hr and the length is changing at 8 ft/hr, so (using algebra):

24 = (2*8) + (2*dw/dt)

Dividing by 2 gives us: 12 = 8 + dw/dt

Subtracting 8 from both sides: 4 = dw/dt

The width of the drainage pond is changing at 4 ft/hr.

*That’s it!*

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