Derivative as a Rate Measure

1. Concept Overview

A derivative represents how fast one quantity changes with respect to another.

If we have a function:

[y=f(x)],

then its derivative:

[\dfrac{d}{dx}\big(f(x)\big)]

represents the rate of change of y with respect to x.

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2. Real-Life Meaning

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3. Important Interpretation Rules

If [\dfrac{dy}{dx} > 0] → y is increasing as x increases
If [\dfrac{dy}{dx} < 0] → y is decreasing
If [\dfrac{dy}{dx} = 0] → y is constant / turning point

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4. Key Mathematical Statement

• If [y=f(x)], then
  [\dfrac{dy}{dx} ] = rate of change of  y w.r.t x
• If a quantity Q depends on t (time):
  [\dfrac{dQ}{dt}] = instantaneous rate of change of Q w.r.t time

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4. A Few Simple Interpretations

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5. Geometrical Meaning

Derivative gives slope of tangent → steepness of graph at a point.

Larger value = steeper curve
Zero value = horizontal tangent

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6. Examples with Step by Step Solution

Example 1 — Kinematics (position → velocity & acceleration)

A particle moves along a line with position [s(t)=4t^{3}-3t^{2}] (metres) where [t] is time in seconds. Find (a) the velocity [\dfrac{ds}{dt}] and (b) the acceleration [\dfrac{d^{2}s}{dt^{2}}] at [t=2\ \text{s}].

Step-by-step Solution:

1. Velocity is the time derivative of position:
   [\dfrac{ds}{dt}][=\dfrac{d}{dt}\big(4t^{3}-3t^{2}\big)=12t^{2}-6t].
2. Evaluate at [t=2]:
   [\dfrac{ds}{dt}\Big|_{t=2}][=12(2)^{2}-6(2)][=12\cdot 4 -12][ =48-12=36\ \text{m/s}].
3. Acceleration is derivative of velocity:
   [\dfrac{d^{2}s}{dt^{2}}][=\dfrac{d}{dt}(12t^{2}-6t)=24t-6].
4. Evaluate at [t=2]:
   [\dfrac{d^{2}s}{dt^{2}}\Big|_{t=2}][=24(2)-6=48-6=42\ \text{m/s}^{2}].

Conclusion:
[\dfrac{ds}{dt}\big|_{t=2}][=36\ \text{m/s},][\quad][ \dfrac{d^{2}s}{dt^{2}}\big|_{t=2}][=42\ \text{m/s}^{2}.]

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Example 2 — Geometry (area growth of a circle)

The radius of a circle grows at a constant rate [\dfrac{dr}{dt}=0.05\ \text{cm/s}]. Find the rate of change of the area when [r=10\ \text{cm}]. (Area [A=\pi r^{2}].)

Step-by-step Solution:

1. Differentiate area w.r.t time using chain rule:
   [\dfrac{dA}{dt}][=\dfrac{d}{dt}\big(\pi r^{2}\big)][=\pi\cdot 2r\cdot \dfrac{dr}{dt}][ = 2\pi r,\dfrac{dr}{dt}].
2. Substitute [r=10] and [\dfrac{dr}{dt}=0.05]:
   [\dfrac{dA}{dt}][=2\pi(10)\cdot 0.05][ = 20\pi\cdot 0.05][ = \pi].
3. Units: [\text{cm}^{2}/\text{s}].

Conclusion:
[\dfrac{dA}{dt}][=\pi\ \text{cm}^{2}/\text{s}] (i.e., about [3.1416\ \text{cm}^{2}/\text{s}]).

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Example 3 — Projectile (max height & velocity at a time)

A ball is thrown vertically: height [h(t)=20t-5t^{2}] metres. (a) Find the time when the ball reaches maximum height. (b) Find velocity at [t=3\ \text{s}].

Step-by-step Solution:

1. Velocity: [\dfrac{dh}{dt}=20-10t].
2. Maximum height occurs when velocity = 0:
   [20-10t=0 \Rightarrow t=2\ \text{s}].
3. Maximum height value: [h(2)=20(2)-5(2)^{2}][=40-20=20\ \text{m}].
4. Velocity at [t=3]: substitute in velocity expression:
   [\dfrac{dh}{dt}\big|_{t=3}][=20-10(3)][=20-30][=-10\ \text{m/s}].
   (Negative sign → moving downward at 10 m/s.)

Conclusion:
Maximum at [t=2\ \text{s}] with height [20\ \text{m}]; velocity at [t=3] is [-10\ \text{m/s}].

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Example 4 — Geometry (water filling a conical tank — related rates)

Water fills a right circular conical tank with fixed dimensions: cone height [H=12\ \text{cm}] and base radius [R=6\ \text{cm}]. Let water depth be [h(t)] (cm). Given [\dfrac{dh}{dt}=0.2\ \text{cm/s}] when [h=4\ \text{cm}], find [\dfrac{dV}{dt}] at that instant. (Volume of cone: [V=\dfrac{1}{3}\pi r^{2}h].)

Step-by-step Solution:

1. Similar triangles give [\dfrac{r}{h}][=\dfrac{R}{H}][=\dfrac{6}{12}][=\dfrac{1}{2}], so [r=\dfrac{h}{2}].
2. Substitute into volume:
   [V][=\dfrac{1}{3}\pi\big(\dfrac{h}{2}\big)^{2}h ][= \dfrac{1}{3}\pi\cdot \dfrac{h^{2}}{4}\cdot h ][= \dfrac{1}{12}\pi h^{3}].
3. Differentiate w.r.t time:
   [\dfrac{dV}{dt}][=\dfrac{1}{12}\pi\cdot 3h^{2}\dfrac{dh}{dt} ][= \dfrac{1}{4}\pi h^{2}\dfrac{dh}{dt}].
4. Substitute [h=4] and [\dfrac{dh}{dt}=0.2]:
   [\dfrac{dV}{dt}][=\dfrac{1}{4}\pi (4)^{2}\cdot 0.2 ][= \dfrac{1}{4}\pi \cdot 16 \cdot 0.2 ][= 4\pi\cdot 0.2 ][= 0.8\pi].
5. Units: [\text{cm}^{3}/\text{s}].

Conclusion:
[\dfrac{dV}{dt}][=0.8\pi\ \text{cm}^{3}/\text{s}] (≈ [2.513\ \text{cm}^{3}/\text{s}]).

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Example 5 — Geometry + Kinematics (sliding ladder — related rates)

A ladder of length [13\ \text{m}] rests against a vertical wall. The foot slides away from the wall at [\dfrac{dx}{dt}=1.5\ \text{m/s}]. Find the rate at which the top of the ladder descends [\dfrac{dy}{dt}] when the foot is [x=5\ \text{m}]. (Relation: [x^{2}+y^{2}=13^{2}].)

Step-by-step Solution:

1. Differentiate [x^{2}+y^{2}=169] w.r.t time [t]:
   [2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} ][= 0].
2. Solve for [\dfrac{dy}{dt}]:
   [\dfrac{dy}{dt} = -,\dfrac{x}{y}\dfrac{dx}{dt}].
3. When [x=5], compute [y]:
   [y][=\sqrt{169 – x^{2}}][=\sqrt{169-25}][=\sqrt{144}][=12\ \text{m}].
4. Substitute values: [\dfrac{dy}{dt} ][= -\dfrac{5}{12}\cdot 1.5 ][= -\dfrac{7.5}{12} ][= -0.625\ \text{m/s}].
   (Negative → top is descending.)

Conclusion:
[\dfrac{dy}{dt} ][= -0.625\ \text{m/s}] when [x=5\ \text{m}]. The top descends at [0.625\ \text{m/s}].

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6. Conceptual Questions with Solutions

1. What does [\dfrac{ds}{dt}] physically represent?

It represents the **rate of change of displacement with respect to time**, i.e., **velocity**.

2. If [\dfrac{dy}{dx} = 0], what does it mean?

The function has **no change at that point**; slope of tangent is zero → turning point.

3. If [\dfrac{dy}{dx} > 0], what does it mean?

Function **increases** as x increases.

4. If [\dfrac{dV}{dt}] is positive, what does it indicate?

The volume is **increasing over time**.

5. What does [\dfrac{dA}{dr}] mean where [A=\pi r^{2}]?

Rate of change of area with respect to radius.

6. In physics, what does [\dfrac{dv}{dt}] represent?

**Acceleration**.

7. Why do we use derivatives to express speed?

Because speed measures **how fast position changes** with time.

8. If [\dfrac{dy}{dx}] is negative, how does the graph look locally?

The graph slopes **downward** from left to right.

9. State the derivative meaning if y relates to time?

[\dfrac{dy}{dt}] = instantaneous **speed of change** in y.

10. Give a real-life example of derivative as rate measure.

Fuel consumption rate of a car: [\dfrac{dF}{dt}].

11. What is the meaning of [\dfrac{dP}{dt}] in economics?

Rate of change of **profit** with respect to time.

12. If distance reduces over time, what is sign of [\dfrac{ds}{dt}]?

Negative.

13. If [A] is maximum at some instant, what is [\dfrac{dA}{dt}] there?

Zero (turning point).

14. If [\dfrac{dy}{dx}] increases with x, what does it signify?

The function is rising **faster**.

15. Can rate of change be a constant? Give example.

Yes, e.g., [y=5x], then [\dfrac{dy}{dx}=5], constant.

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7. FAQ / Common Misconceptions

1. Is derivative always with respect to time?

No. It can be w.r.t **any independent variable**.

2. If rate is zero, does the function stop forever?

No, it may change direction or start increasing again.

3. A positive derivative always means increasing forever?

No, only **at that instant**.

4. Does a large derivative mean large function value?

No, derivative is about **change**, not magnitude.

5. Can derivative be fractional?

Yes, any real number.

6. If graph is vertical, derivative exists?

No. Vertical tangent ⇒ derivative undefined.

7. If function is constant, is rate zero?

Yes.

8. Do derivatives apply only in mathematics class?

No. Widely used in physics, biology, economics, etc.

9. Is rate always positive in motion?

No. Motion can be backward → negative velocity.

10. Is a higher derivative meaningless?

No. 2nd derivative = acceleration, 3rd derivative = jerk, etc.