01 · Reverse the connection
y = x³. At x = 2, dy/dt = 36. Find dx/dt.
Hint
36 = 3x²(dx/dt).
Worked solution
dx/dt = 36/12 = 3. Dividing by dy/dx reverses the connection at this input.
Understand · explore · practise
Connected rates of change
Connect rates using the chain rule: radius, area, volume and depth in spheres, hemispheres, cubes, cylinders and cones. Original worked examples, a manual filling model and practice with solutions.
Before you startChain rule; area and volume formulae; similar triangles
01 / Connect the changing quantities
A derivative with respect to length is not yet a rate per second.
dV/dt = (dV/dh)(dh/dt)
If volume depends on depth and depth changes with time, use the chain rule to connect the two rates. Identify the quantity whose time rate is given and the one whose time rate is required.
In the model, select a flow and compare several depths. More water is needed for each extra centimetre of depth near the top.
Same flow, different depth rateExplore
h = 3 cm; r = 1.5 cm; V = 2.25π cm³; dV/dh = 2.25π cm².
dV/dt = 3π cm³/s; dh/dt = 1.33333 cm/s. The water is rising.
Each selection is a separate instant, not an animation of elapsed time. The full tank shown has height 6 cm. The ideal formula applies while 0 < h < 6; at h = 6 the positive-rate result is the limiting rate just before overflow.
02 / Choose the direction of the chain
Write the equation before rearranging it.
y = x² + 1 and dx/dt = 4 at x = 3Worked example
dy/dx = 2x
Differentiate the relation between the quantities.
dy/dt = (dy/dx)(dx/dt) = 2x·4
Connect both changes through x.
At x = 3, dy/dt = 24
Substitute the instant only after differentiation.
02 · Audit the units
If dV/dh is measured in cm² and dh/dt in cm/s, what units does their product have?
Hint
Multiply the units as well as the numerical values.
Worked solution
cm² × cm/s = cm³/s, the correct units for a volume rate.
03 / Connect a sphere’s volume and radius
Differentiate before substituting the radius.
A sphere grows at dV/dt = 16π cm³/s. Find dr/dt when r = 2 cm.Worked example
V = (4/3)πr³, so dV/dr = 4πr²
The derivative is the sphere’s surface area.
16π = 4πr²(dr/dt)
Use the chain rule.
At r = 2, dr/dt = 1 cm/s
This is an instantaneous radius rate.
For A = 4πr², dA/dt = 8πr(dr/dt) = 16π cm²/s
A second connection gives the surface-area rate.
03 · Another sphere
A sphere has volume rate 36π cm³/s. Find its radius rate at r = 3 cm.
Hint
36π = 4π·3²(dr/dt).
Worked solution
dr/dt = 1 cm/s.
04 · Then its area
At the same instant, find the surface-area rate of the sphere in question 03.
Hint
dA/dt = 8πr(dr/dt).
Worked solution
dA/dt = 24π cm²/s.
04 / Use the correct solid and surface
A hemisphere has half the sphere’s volume.
A hemispherical solid shrinks at 18π cm³/s. Find its radius rate when r = 3 cm.Worked example
V = (2/3)πr³; dV/dr = 2πr²
Use a hemisphere, not a full sphere.
dV/dt = −18π
A loss is a negative signed rate.
−18π = 18π(dr/dt), so dr/dt = −1 cm/s
The radius decreases.
Its curved area is A = 2πr², so dA/dt = 4πr(dr/dt) = −12π cm²/s
Including the circular base would instead give total area 3πr².
05 · A hemispherical loss
A hemisphere loses volume at 8π cm³/s. Find dr/dt when r = 4 cm.
Hint
Use −8π = 2π·4²(dr/dt).
Worked solution
dr/dt = −1/4 cm/s. Its radius decreases at 1/4 cm/s.
06 · Curved or total area?
For question 05, find the curved-area rate and the total-area rate including the base.
Hint
Differentiate 2πr² and 3πr² separately.
Worked solution
The curved-area rate is 4π·4·(−1/4) = −4π cm²/s. The total-area rate is 6π·4·(−1/4) = −6π cm²/s.
05 / Connect a cube’s side, volume and area
Name the side length to avoid confusing the quantities.
A cube’s volume increases at 81 cm³/s. Its side length is s = 3 cm.Worked example
V = s³, so dV/dt = 3s²(ds/dt)
Volume depends on the changing side.
81 = 27(ds/dt), so ds/dt = 3 cm/s
This rate applies at the stated side length.
Surface area A = 6s²
All six faces contribute.
dA/dt = 12s(ds/dt) = 108 cm²/s
The surface-area rate is different from the side rate.
07 · Growing cube
A cube grows at 96 cm³/s. Find its side-length rate when the side is 4 cm.
Hint
96 = 3·4²(ds/dt).
Worked solution
ds/dt = 2 cm/s.
08 · Its surface-area rate
Find the surface-area rate for the cube in question 07.
Hint
dA/dt = 12s(ds/dt).
Worked solution
dA/dt = 12·4·2 = 96 cm²/s. The matching numerical value of the volume rate is a coincidence; the units differ.
06 / A fixed cross-section gives a constant depth rate
State which dimensions are fixed.
Water enters a vertical cylinder of fixed radius 3 cm at 18π cm³/s.Worked example
V = π·3²h = 9πh
The tank radius does not change.
18π = 9π(dh/dt)
Connect volume and water depth.
dh/dt = 2 cm/s
This stays constant while the inflow and cross-section stay constant and before the tank overflows.
09 · Change the radius
A cylinder has fixed radius 2 cm and inflow 12π cm³/s. Find its depth rate.
Hint
dV/dh = 4π.
Worked solution
dh/dt = 3 cm/s.
10 · Both dimensions change
For V = πr²h, derive dV/dt if both r and h change with time.
Hint
Use the product rule and chain rule.
Worked solution
dV/dt = 2πrh(dr/dt) + πr²(dh/dt). Dropping the first term is valid only when dr/dt = 0.
07 / Use similarity before differentiating a cone
The water surface radius changes as its depth changes.
A conical tank has height 6 cm and top radius 3 cm.Worked example
r/h = 3/6, so r = h/2
Similar triangles connect the water radius and depth.
V = (1/3)πr²h = πh³/12
Eliminate r so only h remains.
dV/dh = πh²/4
The horizontal water cross-section is π(h/2)².
If dV/dt = 3π, then dh/dt = 12/h²
The depth rate decreases as the depth grows.
At h = 2, dh/dt = 3 cm/s; at h = 4, dh/dt = 3/4 cm/s
Equal volume rates do not mean equal depth rates.
Compare the same volume flow at two depths
Pause, replay or seek freely. The notes explain the same idea and stay in view.
11 · Different cone dimensions
A cone has height 12 cm and top radius 4 cm. Water enters at 9π cm³/s. Find dh/dt when h = 3 cm.
Hint
r = h/3, so V = πh³/27.
Worked solution
dV/dh = πh²/9. At h = 3, 9π = π(dh/dt), giving dh/dt = 9 cm/s.
08 / Keep signed rates separate from positive speeds
A rate of decrease is often reported as a positive magnitude.
The model cone drains at 3π cm³/s when h = 2 cm.Worked example
dV/dt = −3π
The phrase “drains at” supplies the negative sign.
−3π = (πh²/4)(dh/dt)
The geometric derivative stays positive.
dh/dt = −3 cm/s
The signed depth rate is negative.
The depth decreases at 3 cm/s
The positive magnitude describes the speed of decrease.
12 · Drain at another depth
The model cone drains at 6π cm³/s when h = 4 cm. Find the signed depth rate and the speed of decrease.
Hint
−6π = 4π(dh/dt).
Worked solution
dh/dt = −3/2 cm/s. The depth decreases at 3/2 cm/s.
09 / Use a rate that depends on the current size
Do not treat a changing rate as a constant.
A sphere has volume rate dV/dt = 2r cm³/s when r is measured in cm.Worked example
4πr²(dr/dt) = 2r
Use the rate at the current radius.
dr/dt = 1/(2πr), for r > 0
Cancel only a nonzero radius.
At r = 5, dr/dt = 1/(10π) cm/s
The coefficient 2 carries units cm²/s in this model.
13 · Constant surface-area rate
A sphere’s surface area increases at 20π cm²/s. Find its volume rate when r = 5 cm.
Hint
First find dr/dt from A = 4πr².
Worked solution
dr/dt = 20π/(8π·5) = 1/2 cm/s. Then dV/dt = 4π·5²·(1/2) = 50π cm³/s.
10 / Substitute the state after differentiating
A numerical value at one instant is not a constant identity.
Differentiate V = (4/3)πr³ first. Then set r = 2.
Replacing r with 2 at the start gives V = 32π/3, a number describing one instant. Differentiating that number as though it described every time loses the changing radius.
If a problem gives the volume instead of the radius, first recover the radius using the original geometry, then evaluate the already-derived rate relation.
14 · Recover the missing radius
A sphere has volume 36π cm³ and volume rate 18π cm³/s. Find its radius rate at that instant.
Hint
(4/3)πr³ = 36π gives r = 3.
Worked solution
dr/dt = 18π/(4π·3²) = 1/2 cm/s.
11 / Interpret the model’s limits
A formula has a physical range as well as an algebraic domain.
For the cone model, h must be positive. The ideal expression dh/dt = 12/h² becomes unbounded as h approaches zero from above. It does not predict a finite initial depth rate at the empty apex. Real inflow, water motion and the shape of the tip may need a more detailed model.
At the top of a full tank, further incoming water can overflow. The formula for a rising contained depth no longer applies beyond the tank’s height. Check the geometry and state the range you use.
15 · Compare depths without recalculating
For a cone with fixed positive inflow, what happens to dh/dt when the water depth doubles while remaining within the tank?
Hint
The rate is proportional to 1/h².
Worked solution
It becomes one quarter of its previous value. The horizontal cross-sectional area becomes four times as large.
12 / Use a reliable connected-rates method
Geometry, derivative, signed rate, instant, units.
- Define the changing dimensions and write the geometric relationship.
- Use similarity or another constraint to reduce the variables if necessary.
- Differentiate before substituting a single instant.
- Connect the derivatives with the chain rule and keep the sign of the given rate.
- Substitute the stated size, rearrange and include units.
- Interpret the sign and check the model’s physical range.
16 · Diagnose a wrong chain
A student uses dr/dt = (dV/dr)(dV/dt). Explain and correct the error.
Hint
Start from dV/dt = (dV/dr)(dr/dt).
Worked solution
The student multiplied instead of dividing. Where dV/dr ≠ 0, dr/dt = (dV/dt)/(dV/dr). The corrected units are (cm³/s)/cm² = cm/s.
Section 1 of 12 · Connect the changing quantities