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The radius of a balloon grows at 1 cm/s. How fast is the volume…

Step by step, with the reason for every step · Related rates · Calculus

Step by step

The question

Question: The radius of a balloon grows at 1 cm/s.

How fast is the volume growing when the radius is 5 cm?

1 Draw and name

Draw it.

As air goes in, the volume V grows, and so does the radius r.

The radius grows by 1 cm per second: dr/dt = 1.

drdt = 1 cm/s

dr/dt means how fast r changes: the change each second.

drdt: how fast r changes, per second

We want dV/dt at the moment when r = 5.

dVdt = ?

2 Write the equation

The volume of a ball: V = 4/3 π r cubed.

V = 43πr3

Careful!

Do not put r = 5 in yet!

r changes with time.

Numbers go in only after the derivative.

Careful: not V = 43π(5)³ first ✗

3 Take the derivative

Both V and r change with time.

The chain rule: r cubed gives 3r squared times dr/dt.

π waits.

dVdt = 4πr2 drdt

4/3 times 3 is 4.

43 × 3 = 4

4 Put the numbers in

Put in r = 5 and dr/dt = 1.

dVdt = 4π(5)²(1)

5 Solve for the rate

Multiply the numbers: 100π.

= 100π cm³/s

With π ≈ 3.14159, about 314.16 cm³ per second.

≈ 314.16 cm³/s

6 Check

Let's check: a tiny moment later, 0.001 seconds.

r is 5.001.

The volume grows by about 0.31422.

r = 5.001: V grows by ≈ 0.31422

That is about 314.22 per second.

Right!

change ÷ 0.001 ≈ 314.22

Let's look back.

First: draw and name.

Then: write the equation.

Then: take the derivative.

Then: put the numbers in.

Then: solve for the rate.

Last: check.

Answer
≈ 314.16 cm³/s

Your turn

Try this one yourself, then check:

Show the answer
The volume grows at 48π ≈ 150.80 cm³/s.

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