Q.A swimming pool is to be drained for cleaning. If $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain and $L = 200(10 - t)^2$, how fast is the water running out at the end of $5$ seconds? What is the average rate at which the water flows out during the first $5$ seconds?

Concept: Related Rates — we differentiate $L(t)$ to find the instantaneous rate of change, and use the slope formula for the average rate.

Step 1: Instantaneous rate at $t=5$

$L = 200(10 - t)^2$

Differentiate with respect to $t$:

$\frac{dL}{dt} = 200 \cdot 2(10 - t) \cdot (-1) = -400(10 - t)$

At $t = 5$:

$\frac{dL}{dt} = -400(10 - 5) = -400 \times 5 = -2000$ litres/second.

The negative sign means water is leaving.

Step 2: Average rate over first 5 seconds

Average rate $= \frac{L(5) - L(0)}{5 - 0}$

$L(0) = 200(10)^2 = 200 \times 100 = 20000$ litres …

1. Understanding the problem — what are we really being asked?

We have a pool draining. The volume of water left at time $t$ seconds is given by:

Two different rates are asked for:

• How fast is the water running out at the end of 5 seconds?

  That’s the instantaneous rate of change of $L$ with respect to $t$, evaluated at $t=5$. Since water is leaving, this rate will be negative — but the question asks “how fast”, so we give the positive magnitude.

• What is the average rate at which water flows out during the first 5 seconds?

  That’s the average rate of change of $L$ over the interval $t=0$ to $t=5$. It’s simply the total change in volume divided by the total time.

Both are examples of related rates — we are relating the change in volume to the change in time.

2. Step-by-step solution

Step 1: Find the instantaneous rate of change (derivative)

We have $L(t) = 200(10 - t)^2$. Differentiate with respect to $t$:

The negative sign tells us that $L$ is decreasing — water is leaving.

At $t = 5$ seconds:

So the instantaneous rate at which water is running out at $t=5$ is $2000$ litres per second (the magnitude; the negative indicates direction — outflow).

Step 2: Find the average rate over the first 5 seconds

Average rate of change of $L$ from $t=0$ to $t=5$ is:

$\text{Average rate} = \frac{L(5) - L(0)}{5 - 0}$ …

Method: Distinguishing Instantaneous Rate from Average Rate of Change

When a problem gives a quantity as an explicit function of time, $Q(t)$, and asks both "how fast is it changing at a specific moment" and "what is the average rate over an interval," these are two different calculations from the same function — computing one does not answer the other.

Steps

Step 1: For the instantaneous rate, differentiate $Q(t)$.

This is itself a function of $t$; "the rate at time $t=a$" means evaluate $Q'(a)$.

Step 2: Substitute the specific time into $Q'(t)$.

This single number is the instantaneous answer — its sign tells you whether the quantity is increasing or decreasing at exactly that instant.

Step 3: For the average rate, use the secant-slope formula — never the derivative.

Over an interval $[t_1,t_2]$:

This needs the values of $Q$ at the two endpoints, not $Q'$ at all. …

Common Mistakes

Mistake 1: Dropping the chain-rule factor of $-1$ when differentiating $(10-t)^2$

Why it's wrong: writing $\frac{dL}{dt}=400(10-t)$ instead of $-400(10-t)$ would give a positive rate at $t=5$, incorrectly implying water is flowing INTO the pool rather than draining out. Correct approach: differentiate the inner function $(10-t)$ too, which contributes the extra factor of $-1$.

Mistake 2: Using the same computation for both the instantaneous rate and the average rate

Why it's wrong: the question asks for two genuinely different things — the derivative $\frac{dL}{dt}\big|_{t=5}$ (instantaneous, gives $2000$ L/s) and $\frac{L(5)-L(0)}{5-0}$ (average over the interval, gives $3000$ L/s) — treating them as interchangeable or reporting only one loses marks on the other part. Correct approach: compute both separately using their own correct definitions. …