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Longhand

MATH 100 related rates practice

In a related rates problem two quantities change together, you are told how fast one of them is changing, and you are asked about the other. The calculus is a single line of implicit differentiation with respect to time. Everything difficult happens before and after it.

The reliable procedure is the same every time: name the variables, write a relationship between them that holds at every instant, differentiate both sides with respect to time, and only then substitute the values at the particular instant you are asked about.

That ordering matters more than anything else in the topic. Substituting first freezes a variable that is supposed to be moving, and the equation collapses to nothing useful.

Begin practice sessionBrowse all 4 questions

What you should be able to do

  • Name variables and rates explicitly

    Write down what each letter means and which rate you are given and want. Related rates questions are mostly translation, and an unlabelled diagram is where translations go wrong.

  • Find a relationship that holds at every instant

    Pythagoras for a sliding ladder, a volume formula for a filling tank, similar triangles for a shadow. The relationship must stay true as things move, not just at the moment of interest.

  • Eliminate variables using the fixed geometry

    A cone of fixed shape gives a constant ratio between radius and depth, which lets you write the volume in one variable before differentiating.

  • Differentiate with respect to time

    Every variable produces its own rate through the chain rule. A squared term becomes 2x·dx/dt, and a product of two varying quantities needs the product rule.

  • Substitute the instant last, and interpret the sign

    Put the numbers in only once the differentiation is complete. A negative rate means the quantity is decreasing, and saying so is part of the answer.

Where marks are usually lost

  • Substituting the given instant before differentiating

    Putting x = 3 into x² + y² = 25 first makes x a constant, its derivative zero, and the whole equation useless. Differentiate, then substitute.

  • Treating a changing dimension as fixed

    In a draining cone the radius of the water surface shrinks with the depth. Using the container radius throughout gives a rate that is wrong by a large factor.

  • Answering the wrong rate

    A shadow question may ask how fast the tip moves, not how fast the shadow lengthens. They are different quantities, and both appear among the plausible-looking numbers.

  • Dropping the sign of the given rate

    Water draining out is a negative dV/dt. Entering it as positive produces a level that rises, which the question has already told you it does not.

  • Forgetting the chain rule on a squared term

    Differentiating r² with respect to time gives 2r·dr/dt, not 2r. Without the rate factor there is nothing to solve for.

Practice questions

Every question below is original material written for Longhand. Open a solution to read the full working, or answer it yourself to have it counted towards your progress.

  • Standard3 marks · 5 min

    A 5 m ladder leans against a vertical wall. The foot of the ladder slides away from the wall at 0.6 m/s. How fast is the top of the ladder sliding down when the foot is 3 m from the wall?

    Answer this question yourself
  • Standard3 marks · 5 min

    A spherical balloon is inflated at a constant 100 cm³/s. How fast is its radius increasing at the instant the radius is 5 cm?

    V=43πr3V = \frac{4}{3}\pi r^{3}
    Answer this question yourself
  • Challenging4 marks · 6 min

    Water drains from an inverted right circular cone at 8 cm³/s. The cone is 12 cm tall and 6 cm in radius at the top. How fast is the water level falling when the water is 4 cm deep?

    V=13πr2hV = \frac{1}{3}\pi r^{2} h
    Answer this question yourself
  • Standard3 marks · 5 min

    A 1.8 m tall person walks away from the base of a 5.4 m lamppost at 1.2 m/s along level ground. How fast is the tip of their shadow moving away from the lamppost?

    Answer this question yourself

Common questions

Should I draw a diagram?
Almost always. Labelling the changing lengths on a sketch is how you find the relationship between them, and it makes clear which quantities are fixed and which are moving.
Why can I not just substitute the numbers at the start?
Because the values given describe one instant, not the whole motion. Treating a moving quantity as a constant removes its derivative, and with it the rate you were asked to find.
What units should the answer have?
The units of the quantity divided by time — centimetres per second for a level that is falling, metres per second for a distance. Carrying units through the final line is a cheap check that you differentiated the right thing.

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