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A short class with Sarah and her students. The other ways are under More.
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The area of a circle is A = πr². What does dA/dr tell us when r = 5 cm?
dA/dr = 2πr, which is 10π. So the area grows by about 10π square centimetres for each extra centimetre of radius.
Good. Now ripples on a lake spread so that the radius grows at 5 cm per second. How do we get the rate of change of area with time?
Use the chain rule: dA/dt = dA/dr times dr/dt, which is 2πr times 5.
And when the radius is 8 cm?
2π times 8 times 5, which is 80π square centimetres per second.
Exactly. Notice that we differentiated first and put in r = 8 only at the end.
A 5 m ladder leans on a wall, and its foot moves away at 2 cm/s. Which equation links the foot's distance x and the height y?
Pythagoras: x² + y² = 25.
Differentiate with respect to time and solve for dy/dt.
2x dx/dt + 2y dy/dt = 0, so dy/dt = −(x/y) dx/dt. With x = 4 and y = 3, that is −8/3 cm/s.
What does the minus sign mean?
That the height is decreasing. The top is sliding down at 8/3 centimetres per second.
For s = t³ − 6t² + 9t, find the velocity and the acceleration.
v = 3t² − 12t + 9 and a = 6t − 12.
When is the particle at rest?
When v = 0. Factorising gives 3(t − 1)(t − 3), so at t = 1 second and t = 3 seconds.
Now economics. If C(x) is the total cost of x units, what is C′(x)?
The marginal cost, roughly the extra cost of producing one more unit.
dy/dx measures how much y changes for each unit change in x.
the chain rule gives dy/dt = (dy/dx) × (dx/dt) for linked quantities.
a positive rate means increasing and a negative rate means decreasing; "falling at 3 cm/s" means a rate of −3 cm/s.
put in the values for the given instant only after differentiating the general relation.
v = ds/dt and a = dv/dt; marginal cost is C′(x) and marginal revenue is R′(x).
For a circle, dA/dr = 2πr, so when r = 5 cm the area changes at 10π cm² per cm of radius.
Ripples whose radius grows at 5 cm/s enclose area at dA/dt = 2πr(dr/dt) = 80π cm²/s when r = 8 cm.
For s = t³ − 6t² + 9t, v = 3(t − 1)(t − 3), so the particle is momentarily at rest at t = 1 s and at t = 3 s.
For C(x) = 0.005x³ − 0.02x² + 30x + 5000, the marginal cost at x = 3 is C′(3) = 30.015, about ₹30.02.
Link them first, then differentiate; substitute last, and you'll get the rate.
A balloon seller pumps air in at a steady rate. At first the balloon swells quickly, but as it grows the radius creeps up more and more slowly, because the same air must now spread over a much larger surface.
"Minus means shrinking": whenever a quantity is falling, write its rate with a minus sign.
L.D.S.U. - Link the quantities, Differentiate, Substitute, state Units.
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How do you find dA/dt if you know dA/dr and dr/dt?
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A quantity is decreasing at 4 cm/s. What is its rate of change?
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When should you substitute the given values in a related-rate problem?
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How are velocity and acceleration found from displacement s(t)?
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What does marginal cost measure?
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How fast one quantity changes compared with another, given by a derivative such as dy/dx.
Rates of change of two or more quantities that are linked by an equation, so that one rate can be found from another.
The rule dy/dt = (dy/dx) × (dx/dt), used to connect rates that change with time.
The rate of change of displacement with respect to time, v = ds/dt.
The rate of change of velocity with respect to time, a = dv/dt.
The rate of change of total cost with respect to the number of units produced, C′(x).
The rate of change of total revenue with respect to the number of units sold, R′(x).
Worked examples of the derivative as a rate of change: rates with respect to another variable, related rates that change with time, velocity and acceleration, and marginal cost and revenue.
This lesson works through the main kinds of rate-of-change problems. It starts with the rate at which one quantity changes with another (the area of a circle with its radius), then uses the chain rule for quantities that change with time: spreading ripples, a balloon being inflated and a ladder sliding down a wall. It then finds velocity and acceleration from a displacement formula, and marginal cost and marginal revenue from cost and revenue functions. A step-by-step method, with attention to signs and units, ties the examples together.
If y = f(x), then dy/dx is the rate of change of y with respect to x: roughly, how much y changes for each unit change in x. Often both quantities change with time t. Then the chain rule links their rates: dy/dt = (dy/dx) × (dx/dt). A positive rate means the quantity is increasing; a negative rate means it is decreasing.
Find the rate of change of the area of a circle with respect to its radius when r = 5 cm. Since A = πr², dA/dr = 2πr = 10π cm² per cm when r = 5. Each extra centimetre of radius adds about 10π cm² of area at that size.
A stone dropped into a still lake sends out circular ripples whose radius grows at 5 cm/s. How fast is the enclosed area growing when the radius is 8 cm? From A = πr², dA/dt = 2πr × dr/dt = 2π × 8 × 5 = 80π cm²/s.
Air is pumped into a spherical balloon at 900 cm³/s. How fast is the radius increasing when r = 15 cm? From V = (4/3)πr³, dV/dt = 4πr² × dr/dt, so 900 = 4π × 225 × dr/dt = 900π × dr/dt, giving dr/dt = 1/π cm/s, about 0.32 cm/s. The same inflow makes the radius grow more slowly as the balloon gets bigger.
A ladder 5 m long leans against a wall. Its foot is pulled away from the wall at 2 cm/s. How fast is the top sliding down when the foot is 4 m from the wall? Let x be the distance of the foot from the wall and y the height of the top. Then x² + y² = 25, and differentiating gives 2x(dx/dt) + 2y(dy/dt) = 0, so dy/dt = −(x/y)(dx/dt). When x = 4, y = 3, so dy/dt = −(4/3) × 2 = −8/3 cm/s. The negative sign shows the height is decreasing: the top slides down at 8/3 cm/s. Because x/y has no units, dy/dt has the same units as dx/dt.
A particle moves along a line so that its displacement after t seconds is s = t³ − 6t² + 9t metres. Its velocity is v = ds/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3) m/s, and its acceleration is a = dv/dt = 6t − 12 m/s². At t = 2 s, v = 12 − 24 + 9 = −3 m/s, so the particle is moving back towards its starting point, and a = 0. It is momentarily at rest when v = 0, at t = 1 s and t = 3 s.
If the total cost of producing x units is C(x) = 0.005x³ − 0.02x² + 30x + 5000 rupees, the marginal cost is C′(x) = 0.015x² − 0.04x + 30. When 3 units are produced, C′(3) = 0.135 − 0.12 + 30 = 30.015, about ₹30.02: making one more unit adds roughly that much to the total cost. Similarly, if the revenue from selling x units is R(x) = 3x² + 36x + 5 rupees, the marginal revenue is R′(x) = 6x + 36, which is ₹66 when x = 5.