According to a psychologist, the ability of a person to understand spatial concepts is given by $A = \dfrac{1}{3}\sqrt{t}$, where $t$ is the age in years, $t \in [5, 18]$. Show that the rate of increase of the ability to understand spatial concepts decreases with age in between 5 and 18.

According to a psychologist, the ability of a person to understand spatial concepts is given by A = (1)/(3)√(t

Question

A = \dfrac{1}{3}\sqrt{t}

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✓ Free question

Concept understanding — Rate Of Change

Rate of Change

The very first application of the derivative is to measure how fast one quantity changes when another changes. If $y = f(x)$ , then the derivative

\frac{dy}{dx} = f'(x)

is the rate of change of $y$ with respect to $x$ . Geometrically it is the slope of the tangent; physically it tells you how sensitive $y$ is to a small change in $x$ at that instant.

Average versus instantaneous rate

Over an interval from $x$ to $x + h$ , the average rate of change is

\frac{f(x+h) - f(x)}{h}.

As $h \to 0$ this becomes the instantaneous rate $\dfrac{dy}{dx}$ . So the derivative is the limit of the average rate — the rate "right now" rather than "over a stretch".

Rates with respect to time

Many problems track how a quantity changes as time passes. If a quantity $Q$ depends on time $t$ , then $\dfrac{dQ}{dt}$ is its rate of change per unit time. For example, if the radius of a circle is $r$ , its area is $A = \pi r^2$ , and the rate at which the area grows is

\frac{dA}{dt} = 2\pi r \,\frac{dr}{dt}.

Here the chain rule links the rate of change of the area to the rate of change of the radius. A positive derivative means the quantity is increasing; a negative one means it is decreasing.

Tip

When two related quantities both change with time, differentiate the equation connecting them with respect to $t$ . Every variable contributes its own rate, tied together by the chain rule.

Reading the sign and size

$\dfrac{dy}{dx} > 0$ : $y$ increases as $x$ increases.
$\dfrac{dy}{dx} < 0$ : $y$ decreases as $x$ increases.
A large magnitude means a steep, fast change; a value near zero means $y$ is barely responding.

Note

The units matter. If $s$ is in metres and $t$ in seconds, then $\dfrac{ds}{dt}$ is a speed in metres per second. Always attach the right units to a rate — it turns an abstract derivative into a meaningful physical statement.

Everything else in this chapter — tangents, increasing/decreasing behaviour, maxima and minima — builds on this single idea: the derivative is a rate of change.

Rate of change as an application of derivatives is one of the very first topics in the NCERT Class 12 Application of Derivatives chapter, tested in nearly every CBSE board paper and JEE Main sitting. "Rate of change formula class 12 examples" is a top search term, and related-rates problems built on this idea (like the growing-circle example) are a recurring board exam question type.

The ability $A = \frac{1}{3}\sqrt{t}$ increases with age, but its rate of increase (the derivative $dA/dt$ ) is a decreasing function of $t$ for $t \in [5,18]$ , meaning improvement slows down as the child grows older.

Why this question is about the second derivative

When a problem asks you to show that a rate of increase decreases, it is asking about the behaviour of the first derivative itself. The rate of increase of ability is $dA/dt$ . To show that this rate decreases with age, we need to check whether $dA/dt$ is a decreasing function of $t$ . That is a job for the second derivative: if the derivative of $dA/dt$ (i.e. $d^2A/dt^2$ ) is negative over the interval, then the rate of increase is indeed falling.

So the plan is simple: differentiate $A$ once to get the rate of increase, then differentiate again to see its trend.

Step-by-step

1. Write down the given function

A(t) = \frac{1}{3} \sqrt{t} = \frac{1}{3} t^{1/2}, \quad t \in [5, 18]

2. Find the rate of increase — the first derivative

Using the power rule $\frac{d}{dt} t^{n} = n t^{n-1}$ :

\frac{dA}{dt} = \frac{1}{3} \cdot \frac{1}{2} t^{-1/2} = \frac{1}{6} t^{-1/2}

So the rate of increase is

\frac{dA}{dt} = \frac{1}{6\sqrt{t}}

This is positive for all $t > 0$ , so ability is always increasing between ages 5 and 18 — that much is obvious.

3. Find how this rate itself changes — the second derivative

Differentiate $\frac{dA}{dt}$ with respect to $t$ :

\frac{d^2A}{dt^2} = \frac{d}{dt} \left( \frac{1}{6} t^{-1/2} \right) = \frac{1}{6} \cdot \left( -\frac{1}{2} \right) t^{-3/2} = -\frac{1}{12} t^{-3/2}

That is:

\frac{d^2A}{dt^2} = -\frac{1}{12\, t^{3/2}}

4. Interpret the sign

For any $t > 0$ , the denominator $12\,t^{3/2}$ is positive. The negative sign in front makes the second derivative negative for all $t$ in $[5, 18]$ .

Important

A negative second derivative means the first derivative is decreasing. Since the first derivative $dA/dt$ is the rate of increase of ability, a decreasing first derivative means the rate of increase itself falls as age increases.

5. Conclude the behaviour

At age 5, the rate of increase is $\frac{1}{6\sqrt{5}} \approx 0.0745$ units per year. At age 18, it is $\frac{1}{6\sqrt{18}} \approx 0.0393$ units per year — roughly half. The improvement in spatial ability slows down steadily over this age range.

Watch out

A common mistake is to think that because $A$ itself increases, the rate of increase must also increase. That is false: a function can rise at a falling rate (concave down). Always check the second derivative, not just the first.

Tip

You don't need to compute numeric values. The algebraic sign of $d^2A/dt^2$ alone is sufficient: it is negative for all $t > 0$ , so the rate of increase decreases throughout $[5, 18]$ .

✓Final answer

\frac{dA}{dt} = \frac{1}{6\sqrt{t}}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 24 on this concept.

CBSE 2026Set 65/3/11 markMCQ
$5$
›Reveal solutionSolution
We need to find the rate of change of the sphere's volume with respect to its diameter, which is $\frac{dV}{dD}$ . Using the chain rule, $\frac{dV}{dD} = \frac{dV}{dr} \cdot \frac{dr}{dD}$ , we find this rate to be $2\pi r^2$ . For a radius of $5$ cm, the rate is $\boxed{50\pi \text{ cm}^3/\text{cm}}$ .

When we talk about the "rate of change of the volume of a sphere with respect to its diameter," we are essentially asking for the derivative of the volume ( $V$ ) with respect to the diameter ( $D$ ). In mathematical terms, this is $\frac{dV}{dD}$ .

The volume of a sphere is typically expressed in terms of its radius, $r$ . The diameter, $D$ , is related to the radius by $D = 2r$ . To find $\frac{dV}{dD}$ , we can either express $V$ entirely in terms of $D$ and then differentiate, or we can use the chain rule. The chain rule is often more intuitive for problems like this, as it breaks down the problem into smaller, more manageable derivatives. It states that if $V$ depends on $r$ , and $r$ depends on $D$ , then $\frac{dV}{dD} = \frac{dV}{dr} \cdot \frac{dr}{dD}$ .

Let's work through the problem step-by-step.

Identify the relevant formulas and relationships. The volume of a sphere is given by:

$V = \frac{4}{3}\pi r^3$
The relationship between the radius ($r$) and the diameter ($D$) is:
$D = 2r$
From this, we can express $r$ in terms of $D$:
$r = \frac{D}{2}$
We are given that the radius $r = 5$ cm. We need to find $\frac{dV}{dD}$ at this specific radius.

2. Find the rate of change of volume with respect to radius ( $\frac{dV}{dr}$ ).

We differentiate the volume formula $V = \frac{4}{3}\pi r^3$ with respect to $r$ :
$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)$ $\frac{dV}{dr} = \frac{4}{3}\pi \cdot 3r^2$ $\frac{dV}{dr} = 4\pi r^2$
This tells us how quickly the volume changes as the radius changes.

3. Find the rate of change of radius with respect to diameter ( $\frac{dr}{dD}$ ).

We use the relationship $r = \frac{D}{2}$ and differentiate it with respect to $D$ :
$\frac{dr}{dD} = \frac{d}{dD}\left(\frac{D}{2}\right)$ $\frac{dr}{dD} = \frac{1}{2}$
This makes sense: for every unit increase in diameter, the radius increases by half a unit.

4. Apply the Chain Rule to find $\frac{dV}{dD}$ .

Now we combine the two rates using the chain rule:

> [!FORMULA]

> The Chain Rule for related rates:

> $\frac{dV}{dD} = \frac{dV}{dr} \cdot \frac{dr}{dD}$

Substitute the expressions we found in steps 2 and 3:
$\frac{dV}{dD} = (4\pi r^2) \cdot \left(\frac{1}{2}\right)$ $\frac{dV}{dD} = 2\pi r^2$
This expression gives the rate of change of the volume with respect to the diameter for any given radius $r$.

5. Substitute the given radius value.

We are asked to find this rate when the radius is $5$ cm. Substitute $r = 5$ into the expression for $\frac{dV}{dD}$ :
$\frac{dV}{dD} = 2\pi (5)^2$ $\frac{dV}{dD} = 2\pi (25)$ $\frac{dV}{dD} = 50\pi$
The units for volume are $\text{cm}^3$ and for diameter are $\text{cm}$, so the unit for the rate of change is $\text{cm}^3/\text{cm}$.

✓Final answer
$5$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The rate of change of the area of a circle with respect to its radius r at r = 5 cm is$
(a) $12\pi$
(b) $8\pi$
(c) $5\pi$
(d) $10\pi$
›Reveal solutionSolution

The area of a circle is $A=\pi r^2$ ; its rate of change with $r$ is $dA/dr=2\pi r$ .

$A=\pi r^2 \Rightarrow \dfrac{dA}{dr}=2\pi r$ .

At $r=5\text{ cm}$ : $\dfrac{dA}{dr}=2\pi(5)=10\pi$ cm²/cm.

✓Final answer
$10\pi$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The rate of change of area of circle w.r.t. its radius 'r' when r = 4 cm is$
(a) 8π cm²/cm
(b) 6π cm²/cm
(c) 4π cm²/cm
(d) 2π cm²/cm
›Reveal solutionSolution

The rate of change of area with radius is $dA/dr = 2\pi r$ ; substitute $r=4$ .

Area of a circle: $A = \pi r^2$ .

$\frac{dA}{dr} = 2\pi r$

At $r = 4\ \text{cm}$ :

$\frac{dA}{dr}\Big|_{r=4} = 2\pi(4) = 8\pi\ \text{cm}^2/\text{cm}.$

✓Final answer
$8\pi\ \text{cm}^2/\text{cm}$
CBSE 2026Set ANNUAL1 markMCQ
$r$
(a) $10\pi$ cm
(b) $12\pi$ cm
(c) $8\pi$ cm
(d) $11\pi$ cm
›Reveal solutionSolution

Differentiate the area formula $A=\pi r^2$ with respect to $r$ and evaluate at $r=6$ .

Area of a circle: $A = \pi r^2$

$\frac{dA}{dr} = 2\pi r$

At $r=6$ cm:

$\frac{dA}{dr}\bigg|_{r=6} = 2\pi(6) = 12\pi$

✓Final answer
$12\pi$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The rate of change of the area of a circle with respect to its radius r at r=6 cm is:$
(a)
(i) 8\pi
(b)
(ii) 10\pi
(c)
(iii) 11\pi
(d)
(iv) 12\pi
›Reveal solutionSolution
$\dfrac{dA}{dr}=2\pi r = 12\pi$ at $r=6$ cm — option (iv).

Concept. The rate of change of a quantity with respect to a variable is its derivative with respect to that variable.

Steps.

Area of a circle: $A=\pi r^2$ .

Differentiate with respect to $r$ : $\dfrac{dA}{dr}=2\pi r$ .

Substitute $r=6$ cm: $\dfrac{dA}{dr}=2\pi(6)=12\pi$ .

✓Final answer
$\dfrac{dA}{dr}\big|_{r=6}=12\pi\ \text{cm}^2/\text{cm}$
CBSE 2025Set A1 mark
$r = 6$
›Reveal solutionSolution

Area of a circle is $A = \pi r^2$ ; differentiate with respect to $r$ and substitute $r=6$ .

Area of a circle: $A = \pi r^2$ .

$\dfrac{dA}{dr} = 2\pi r$

At $r = 6$ cm:

$\dfrac{dA}{dr}\bigg|_{r=6} = 2\pi(6) = 12\pi \text{ cm}^2/\text{cm}$

✓Final answer
$12\pi$
CBSE 2025Set ANNUAL1 mark
$r$
›Reveal solutionSolution

The rate of change of the area of a circle with respect to its radius is $\dfrac{dA}{dr}$ .

$A=\pi r^2 \;\Rightarrow\; \dfrac{dA}{dr}=2\pi r$

At $r=2.5$ : $\dfrac{dA}{dr}=2\pi(2.5)=5\pi$ cm $^2$ per cm.

✓Final answer
$5\pi$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The rate of change of area of a circle w.r.t. its radius 'r' when r = 5 cm is :$
(a) 8π cm²/cm
(b) 10π cm²/cm
(c) 9π cm²/cm
(d) 16π cm²/cm
›Reveal solutionSolution

Area of a circle $A=\pi r^2$ ; rate of change $\dfrac{dA}{dr}=2\pi r$ , which at $r=5$ gives $10\pi$ .

Given $A=\pi r^2$ , differentiate with respect to $r$ :

$\frac{dA}{dr} = 2\pi r.$

At $r=5\ \text{cm}$ :

$\frac{dA}{dr}\bigg|_{r=5} = 2\pi(5) = 10\pi\ \text{cm}^2/\text{cm}.$

✓Final answer
$10\pi\ \text{cm}^2/\text{cm}$
CBSE 2025Set ANNUAL1 markMCQ
$r=6$
(a) $6\pi\ cm^2/cm$
(b) $6\pi\ cm/cm^2$
(c) $12\pi\ cm^2/cm$
(d) $12\pi\ cm/cm^2$
›Reveal solutionSolution

Differentiate the area formula $A=\pi r^2$ w.r.t. $r$ and substitute $r=6$ .

Area of circle: $A = \pi r^2$ .

$\dfrac{dA}{dr} = 2\pi r$

At $r=6$ cm:

$\dfrac{dA}{dr}\bigg|_{r=6} = 2\pi(6) = 12\pi \text{ cm}^2/\text{cm}$

✓Final answer
$12\pi\ \text{cm}^2/\text{cm}$
CBSE 2024Set 65/3/11 markMCQ
$1.5$
›Reveal solutionSolution
The perimeter of a square is directly proportional to its side length. If the side decreases at $1.5$ cm/s, the perimeter decreases at $4$ times that rate, which is $6$ cm/s. The correct option is (B).

This problem asks us to find the rate at which the perimeter of a square is decreasing, given the rate at which its sides are decreasing. This is a classic "related rates" problem in calculus. The core idea is to establish a relationship between the quantities involved (side length and perimeter), and then differentiate that relationship with respect to time to find how their rates of change are related.

When we talk about a "rate of change," we are essentially talking about a derivative with respect to time. If a quantity is decreasing, its rate of change will be negative.

Here's how we approach it:

Identify the variables and given rates.

Let $s$ be the side length of the square at any given time $t$ .

Let $P$ be the perimeter of the square at any given time $t$ .

We are given that the sides of the square are decreasing at the rate of $1.5$ cm/s. In calculus terms, this means the derivative of the side length with respect to time, $\frac{ds}{dt}$ , is $-1.5$ cm/s. The negative sign indicates a decrease.

Our goal is to find the rate of decrease of the perimeter, which means we need to find $\frac{dP}{dt}$ .

Establish a relationship between the variables.

The formula for the perimeter of a square with side length $s$ is:

$P = 4s$

Differentiate the relationship with respect to time. To find how the rates of change are related, we differentiate both sides of the equation $P = 4s$ with respect to time $t$ . We use the chain rule here.

$\frac{dP}{dt} = \frac{d}{dt}(4s)$
Since $4$ is a constant, we can pull it out of the differentiation:
$\frac{dP}{dt} = 4 \frac{ds}{dt}$
This equation tells us that the rate of change of the perimeter is $4$ times the rate of change of the side. This makes intuitive sense: if each of the four sides shrinks by a certain amount, the total perimeter shrinks by four times that amount.

4. Substitute the given rate and calculate.

We know $\frac{ds}{dt} = -1.5$ cm/s. Substitute this value into the equation from step 3:
$\frac{dP}{dt} = 4(-1.5)$ $\frac{dP}{dt} = -6$
The unit for $\frac{dP}{dt}$ will be cm/s, consistent with the units of $\frac{ds}{dt}$. The result $\frac{dP}{dt} = -6$ cm/s means that the perimeter is changing at a rate of $-6$ cm/s. Since the question asks for the "rate of *decrease*", we state the positive value of this rate. > [!WARNING] > Be careful with the sign. If the problem asks for the "rate of decrease," the answer should be a positive value. The negative sign in $\frac{dP}{dt}$ simply indicates that the quantity is decreasing. Therefore, the rate of decrease of the perimeter is $6$ cm/s. Comparing this with the given options: (A) $1.5$ cm/s (B) $6$ cm/s (C) $3$ cm/s (D) $2.25$ cm/s The calculated rate matches option (B).

✓Final answer
$\boxed{\text{6 cm/s}}$
CBSE 2024Set D1 markMCQ
$r$
(a) $10\pi$
(b) $12\pi$
(c) $8\pi$
(d) $11\pi$
›Reveal solutionSolution

$\frac{dA}{dr}=2\pi r$ , which at $r=6$ equals $12\pi$ .

Area of a circle: $A = \pi r^2$ .

Rate of change of area with respect to radius: $\dfrac{dA}{dr} = 2\pi r$ .

At $r = 6\,\mathrm{cm}$ : $\dfrac{dA}{dr} = 2\pi(6) = 12\pi\ \mathrm{cm}^2/\mathrm{cm}$ .

✓Final answer
$12\pi$
CBSE 2024Set ANNUAL1 mark
$r$
›Reveal solutionSolution

Differentiate the area formula $A=\pi r^2$ w.r.t. $r$ and evaluate at $r=3$ .

$A = \pi r^2 \Rightarrow \dfrac{dA}{dr} = 2\pi r$

At $r = 3$ : $\dfrac{dA}{dr} = 2\pi(3) = 6\pi$ cm² per cm.

✓Final answer
$6\pi$

Why this question is about the second derivative

Step-by-step

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