Rate of Change Math Example 2

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Example 2

medium
Water drains from a tank so that the volume remaining after tt minutes is V(t)=500−20t−t2V(t) = 500 - 20t - t^2 litres (0≤t≤100 \leq t \leq 10). Find the rate at which water is draining at t=3t = 3 minutes.

Solution

  1. 1
    The rate of change of volume is V′(t)V'(t).
  2. 2
    Differentiate: V′(t)=−20−2tV'(t) = -20 - 2t.
  3. 3
    At t=3t = 3: V′(3)=−20−6=−26V'(3) = -20 - 6 = -26 litres per minute.
  4. 4
    The negative sign means volume is decreasing, i.e., water is draining at 26 L/min.

Answer

Water drains at 2626 litres per minute at t=3t = 3.
A negative rate of change means the quantity is decreasing. The magnitude gives the speed of decrease. Always interpret the sign: here V′(3)=−26V'(3) = -26 means the volume is falling at 26 litres per minute.

About Rate of Change

A measure of how quickly one quantity changes with respect to another; the ratio of the change in output to the change in input.

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Example 3 medium

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Example 4 easy

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Example 5 medium

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