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Calculus IApplications

Related rates, one change linked to another.

Draw the picture, connect the quantities, and turn their equation into a rule for their rates. Then practise with shapes, ladders, angles and shadows.

Build the equation · Step 1 of 7

Two quantities change together

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Burgundy: the 15 ft ladder. x is the distance along the ground; y is the height up the wall. Both axes measure ft.
A related-rates problem gives you one rate and asks for another. The bridge is an equation connecting the quantities. A rate such as dxdt{\frac{dx}{dt}} says how fast x changes as time t passes; it is different from x itself.
Try it: A 15 ft ladder's base moves away from a wall at 1 ft/s. At one instant the base is 9 ft away. Which number is a length, and which is a rate?
  1. Draw a picture. Name the changing lengths, areas, volumes or angles.
  2. List what is given and what is asked, including units and signs.
  3. Write an equation that stays true while the objects move.
  4. Differentiate with respect to time, then substitute the values for the requested instant.
  5. Solve for the unknown rate. Explain its sign and include units.

x is the ground distance, y is the height, and t is time in seconds.

Given: dxdt=1  fts{\displaystyle \text{Given: }\frac{dx}{dt}} = {\displaystyle 1\;\frac{\mathrm{ft}}{\mathrm{s}}}
At this instant: x=9  ft{\displaystyle \text{At this instant: }x} = {\displaystyle 9\;\mathrm{ft}}
Wanted: dydt{\displaystyle \text{Wanted: }\frac{dy}{dt}}

The lengths make a right triangle. Its hypotenuse is the fixed ladder.

x2+y2=152{\displaystyle x^2} + {\displaystyle y^2} = {\displaystyle 15^2}
Check yourself
The base is 9 ft away and moves at 1 ft/s. Which is a rate?

Not sure how to turn the story into an equation?

Bring your practice questions. We work through them together until every type feels routine.

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