Sample 2023-24 Class 12 Maths Q38 Solution: Case study: height of a plant y = 4x − x²/2 and its rate of growth

The question

Read the following passage and answer the questions given below:

The relation between the height of the plant (' $y$ ' in cm) with respect to its exposure to the sunlight is governed by the following equation $y = 4x - \dfrac12x^2$ ⁠, where ' $x$ ' is the number of days exposed to the sunlight, for $x \le 3$ ⁠.

(i) Find the rate of growth of the plant with respect to the number of days exposed to the sunlight. (2 marks)

(ii) Does the rate of growth of the plant increase or decrease in the first three days? What will be the height of the plant after 2 days? (2 marks)

Exam answer

y = 4x - \frac12x^2

(i) The rate of growth of the plant with respect to the number of days exposed to sunlight is given by

Why?

"Rate of growth" means how fast the height $y$ changes as the number of days $x$ changes. That is the derivative $\frac{dy}{dx}$ ⁠.

\frac{dy}{dx} = 4 - \frac12(2x)

Why?

Differentiate term by term with the power rule $\frac{d}{dx}x^n = nx^{n - 1}$ ⁠: $4x$ gives 4, and $x^2$ gives $2x$ ⁠, which is then multiplied by $\frac12$ ⁠.

\begin{aligned} \frac{d}{dx}(4x) &= 4 \\ \frac{d}{dx}\left(\frac12x^2\right) &= \frac12(2x) = x \\[4pt] \frac{dy}{dx} &= 4 - x \end{aligned}

$\dfrac{dy}{dx} = 4 - x$

(ii) Let the rate of growth be represented by the function $g(x) = \dfrac{dy}{dx} = 4 - x$ ⁠.

\text{Now, } g'(x) = \frac{d}{dx}\left(\frac{dy}{dx}\right) = -1 < 0

Why?

To tell whether a function increases or decreases, look at the sign of its derivative. Here the function in question is the rate $g(x)$ itself, so we differentiate once more: $\frac{d}{dx}(4 - x) = -1$ ⁠.

A negative derivative means the function decreases.

$\Rightarrow g(x)$ decreases.

So the rate of growth of the plant decreases for the first three days.

Why?

The rate on days 0, 1, 2, 3 is $4, 3, 2, 1$ cm per day: still positive, but smaller each day.

\begin{aligned} g(0) &= 4 - 0 = 4 \\ g(1) &= 4 - 1 = 3 \\ g(2) &= 4 - 2 = 2 \\ g(3) &= 4 - 3 = 1 \end{aligned}

Watch out: the plant is still getting taller, since $\frac{dy}{dx} > 0$ for $x \le 3$ ⁠. It is the speed of growth that decreases.

y(2) = 4 \times 2 - \frac12(2)^2 = 6

Why?

The height after 2 days is the value of $y$ at $x = 2$ ⁠, not the rate: $4 \times 2 = 8$ and $\frac12 \times 4 = 2$ ⁠, so $8 - 2 = 6$ ⁠. The unit is cm.

\begin{aligned} y(2) &= 4(2) - \frac12(2)^2 \\ &= 8 - \frac12(4) \\ &= 8 - 2 = 6 \end{aligned}

The rate of growth decreases, and the height of the plant after 2 days is 6 cm.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: growth of a plant.

Question type: (i) Rate of change with respect to a length; (ii) Show that a function is increasing everywhere

How to spot it: click to see

(i) Type 1. Rate of change with respect to a length

Spot it: "the rate of change of the area (or volume) with respect to the radius". No time is mentioned.

Here: The rate of growth with respect to the number of days is $\frac{dy}{dx}$ ⁠: differentiate $y = 4x - \frac12x^2$ with respect to $x$ to get $4 - x$ cm per day.

(ii) Type 4. Show that a function is increasing everywhere

Spot it: "Show that (or prove that) $f$ is increasing on $\R$ ⁠."

Here: The function to test is the rate $g(x) = 4 - x$ itself. Its derivative $g'(x) = -1$ is negative for every $x$ ⁠, so the rate of growth decreases.

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Q38: Case study: height of a plant y=4x−12x2y = 4x - \frac12x^2y=4x−21​x2 and its rate of growth

Q38: Case study: height of a plant $y = 4x - \frac12x^2$ and its rate of growth