Jamb Mathematics - Lesson Notes on Application of Differentiation for UTME Candidate

Application of differentiation | Jamb Mathematics

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Hearken, O diligent scholar, for the hour of reckoning draws nigh! The great trial on the application of variation approaches, and thou must steel thy mind with wisdom, lest the test of numbers confound thee. Gather thy parchments, sharpen thy quill, and let no theorem remain unmastered, for fortune favors the prepared!

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Application of Differentiation? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Application of Differentiation together and move one step closer to achieving your exam success! Blissful learning.

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Calculation problems involving rate of change

Problem 1: Rate of Change of Area of a Circle

The radius of a circle is increasing at a rate of

3

cm/s. Find the rate at which the area of the circle is increasing when the radius is

10

cm.

Solution:

The area of a circle is given by $A = \pi r^2$ .
Differentiate both sides with respect to time $t$ :
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$
Given $\frac{dr}{dt} = 3$ cm/s and $r = 10$ cm, substitute: $\frac{dA}{dt} = 2\pi (10)(3) = 60\pi$ cm²/s.

Problem 2: Rate of Change of Volume of a Sphere

A spherical balloon is being inflated at a rate of

100

cm³/s. Find the rate at which the radius is increasing when the radius is

5

cm.

Solution:

Volume of a sphere: $V = \frac{4}{3} \pi r^3$
Differentiate:
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Given $\frac{dV}{dt} = 100$ , $r = 5$ : $100 = 4\pi (5)^2 \frac{dr}{dt}$
Solve for $\frac{dr}{dt}$ :
$\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}$ cm/s.

Problem 3: Rate of Change of Volume of a Cube

A cube's side length is increasing at a rate of

2

cm/s. Find the rate at which the volume of the cube is increasing when the side length is

4

cm.

Solution:

Volume of a cube: $V = s^3$
Differentiate:
$\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$
Given $\frac{ds}{dt} = 2$ , $s = 4$ : $\frac{dV}{dt} = 3(4)^2 (2) = 96$ cm³/s.

Problem 4: Ladder Sliding Down a Wall

10

m ladder is leaning against a wall. The bottom is moving away from the wall at

1.5

m/s. How fast is the top of the ladder sliding down when the bottom is

6

m from the wall?

Solution:

Let $x$ be the distance from the wall, $y$ be the height of the ladder:
$x^2 + y^2 = 100$
Differentiate:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
Given $x = 6$ , solve for $y$ :
$y = \sqrt{100 - 36} = 8$
Solve for $\frac{dy}{dt}$ : $2(6)(1.5) + 2(8) \frac{dy}{dt} = 0$ $18 + 16 \frac{dy}{dt} = 0$ $\frac{dy}{dt} = -\frac{18}{16} = -1.125$ m/s.

Problem 5: Water Draining from a Conical Tank

A water tank in the shape of an inverted cone has a height of

10

m and a base radius of

4

m. If water is leaking at

3

m³/min, find the rate at which the water level is decreasing when the water is

6

m deep.

Solution:

Volume of a cone: $V = \frac{1}{3} \pi r^2 h$
Express $r$ in terms of $h$ using similar triangles:
$\frac{r}{h} = \frac{4}{10} \Rightarrow r = \frac{2}{5} h$
Substituting into volume formula:
$V = \frac{1}{3} \pi \left( \frac{2}{5} h \right)^2 h = \frac{4}{75} \pi h^3$
Differentiate:
$\frac{dV}{dt} = \frac{4}{25} \pi h^2 \frac{dh}{dt}$
Given $\frac{dV}{dt} = -3$ , $h = 6$ : $-3 = \frac{4}{25} \pi (6)^2 \frac{dh}{dt}$ $-3 = \frac{144}{25} \pi \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{-3 \times 25}{144\pi} = \frac{-75}{144\pi} \approx -0.165$ m/min.

Problem 6: Expanding Square

The side length of a square is increasing at a rate of

4

cm/s. Find the rate at which its perimeter is increasing when the side length is

10

cm.

Solution:

Perimeter of a square: $P = 4s$
Differentiate: $\frac{dP}{dt} = 4 \frac{ds}{dt}$
Given $\frac{ds}{dt} = 4$ , $s = 10$ : $\frac{dP}{dt} = 4(4) = 16$ cm/s.

Problem 7: Expanding Rectangle

A rectangle’s length is increasing at

3

cm/s, and its width is increasing at

2

cm/s. Find the rate at which its area is increasing when the length is

8

cm and the width is

5

cm.

Solution:

Area of a rectangle: $A = lw$
Differentiate: $\frac{dA}{dt} = l \frac{dw}{dt} + w \frac{dl}{dt}$
Given $\frac{dl}{dt} = 3$ , $\frac{dw}{dt} = 2$ , $l = 8$ , $w = 5$ : $\frac{dA}{dt} = 8(2) + 5(3) = 16 + 15 = 31$ cm²/s.

Problem 8: Shrinking Circular Pool

The radius of a circular pool is decreasing at a rate of

0.5

m/min. Find the rate at which the circumference is decreasing when the radius is

12

Solution:

Circumference of a circle: $C = 2\pi r$
Differentiate: $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$
Given $\frac{dr}{dt} = -0.5$ , $r = 12$ : $\frac{dC}{dt} = 2\pi (-0.5) = -\pi$ m/min.

Problem 9: Draining Cylindrical Tank

A cylindrical tank with radius

5

m is being emptied at a rate of

10

m³/min. Find the rate at which the height of the water is decreasing.

Solution:

Volume of a cylinder: $V = \pi r^2 h$
Differentiate: $\frac{dV}{dt} = \pi (5)^2 \frac{dh}{dt}$
Given $\frac{dV}{dt} = -10$ : $-10 = 25\pi \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{-10}{25\pi} = \frac{-2}{5\pi}$ m/min.

Problem 10: Moving Particle

A particle moves along the

x

-axis, and its position at time

t

is given by

x = 3t^2 + 2t

. Find the velocity and acceleration at

t = 4

Solution:

Velocity: $v = \frac{dx}{dt} = 6t + 2$ $v = \frac{d x}{d t} = 6 t + 2$
- At $t = 4$ : $v = 6(4) + 2 = 26$ m/s.
Acceleration: $a = \frac{dv}{dt} = 6$ $a = \frac{d v}{d t} = 6$
- At $t = 4$ : $a = 6$ m/s².

Problem 11: Expanding Cylinder

A cylindrical tank with a constant height of

10

m is being filled at

20

m³/min. Find the rate at which the radius is increasing when the radius is

3

Solution:

Volume: $V = \pi r^2 h$ $V = π r^{2} h$ , where $h = 10$ $h = 10$
- $V = 10\pi r^2$
Differentiate: $\frac{dV}{dt} = 20\pi r \frac{dr}{dt}$
Given $\frac{dV}{dt} = 20$ , $r = 3$ : $20 = 20\pi (3) \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{20}{60\pi} = \frac{1}{3\pi}$ m/min.

Problem 12: Melting Ice Sphere

An ice sphere is melting at a rate of

4

cm³/min. Find the rate at which the radius is decreasing when the radius is

6

cm.

Solution:

Volume: $V = \frac{4}{3} \pi r^3$
Differentiate: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Given $\frac{dV}{dt} = -4$ , $r = 6$ : $-4 = 4\pi (6)^2 \frac{dr}{dt}$ $-4 = 144\pi \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{-4}{144\pi} = \frac{-1}{36\pi}$ cm/min.

Problem 13: Changing Angle of Elevation

A plane is flying horizontally at

500

km/h at an altitude of

3

km. Find the rate at which the angle of elevation is changing when the plane is

4

km away.

Solution:

Use $\tan \theta = \frac{3}{x}$ .
Differentiate: $\sec^2 \theta \frac{d\theta}{dt} = -\frac{3}{x^2} \frac{dx}{dt}$ .
Given $x = 4$ , $\frac{dx}{dt} = 500$ , find $\theta$ : $\tan \theta = \frac{3}{4} \Rightarrow \sec^2 \theta = 1 + \tan^2 \theta = \frac{25}{16}$ .
Solve for $\frac{d\theta}{dt}$ : $\frac{25}{16} \frac{d\theta}{dt} = -\frac{3}{16} (500)$ $\frac{d\theta}{dt} = -\frac{1500}{400} = -3.75$ rad/h.

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Calculation problems involving maxima and minima in differentiation

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Problem 1: Maximum Area of a Rectangular Enclosure

A farmer wants to enclose a rectangular field using 200 meters of fencing. What dimensions will maximize the area of the field?

Solution:

Let the length be $x$ and the width be $y$ . The perimeter constraint is: $2x + 2y = 200 \Rightarrow y = 100 - x$ .
Area: $A = xy = x(100 - x) = 100x - x^2$ .
Differentiate: $\frac{dA}{dx} = 100 - 2x$ .
Set $\frac{dA}{dx} = 0$ :
$100 - 2x = 0 \Rightarrow x = 50$ .
Second derivative: $\frac{d^2A}{dx^2} = -2 < 0$ , confirming a maximum.
Dimensions: $x = 50$ , $y = 50$ .

Problem 2: Minimum Surface Area of a Cylinder

Find the dimensions of a closed cylindrical can with volume

1000

cm³ that minimizes surface area.

Solution:

Volume: $V = \pi r^2 h = 1000$ .
Solve for $h$ : $h = \frac{1000}{\pi r^2}$ .
Surface area: $S = 2\pi r^2 + 2\pi r h$ .
Substitute $h$ : $S = 2\pi r^2 + \frac{2000}{r}$ .
Differentiate:
$\frac{dS}{dr} = 4\pi r - \frac{2000}{r^2}$ .
Set $\frac{dS}{dr} = 0$ and solve:
$4\pi r = \frac{2000}{r^2} \Rightarrow r^3 = \frac{500}{\pi} \Rightarrow r \approx 5.42$ cm.
Compute $h$ : $h \approx 10.84$ cm.

Problem 3: Maximum Volume of a Box with Square Base

A box with a square base and no top is made from 600 cm² of material. Find the maximum volume.

Solution:

Let base side be $x$ and height be $h$ .
Surface area: $x^2 + 4xh = 600$ .
Solve for $h$ : $h = \frac{600 - x^2}{4x}$ .
Volume: $V = x^2 h = x^2 \frac{600 - x^2}{4x} = \frac{600x - x^3}{4}$ .
Differentiate:
$\frac{dV}{dx} = \frac{600 - 3x^2}{4}$ .
Solve $\frac{dV}{dx} = 0$ :
$600 - 3x^2 = 0 \Rightarrow x^2 = 200 \Rightarrow x \approx 14.14$ cm.
Compute $h$ : $h \approx 7.07$ cm.

Problem 4: Minimum Distance from a Point to a Line

Find the point on the line

y = 2x + 3

that is closest to the point

(4,0)

Solution:

Distance: $D^2 = (x - 4)^2 + (2x + 3 - 0)^2$ .
Expand: $D^2 = (x - 4)^2 + (2x + 3)^2$ .
Differentiate:
$\frac{dD^2}{dx} = 2(x - 4) + 2(2x + 3)(2)$ .
Set $\frac{dD^2}{dx} = 0$ : $2(x - 4) + 4(2x + 3) = 0$ .
Solve for $x$ :
$2x - 8 + 8x + 12 = 0 \Rightarrow 10x + 4 = 0 \Rightarrow x = -0.4$ .
Compute $y = 2(-0.4) + 3 = 2.2$ .

Problem 5: Maximum Profit

A company finds that its revenue is

R = 100x - x^2

and its cost is

C = 20x + 300

. Find the production level

x

that maximizes profit.

Solution:

Profit: $P = R - C = (100x - x^2) - (20x + 300)$ .
Simplify: $P = -x^2 + 80x - 300$ .
Differentiate: $\frac{dP}{dx} = -2x + 80$ .
Set $\frac{dP}{dx} = 0$ : $-2x + 80 = 0 \Rightarrow x = 40$ .
Second derivative: $\frac{d^2P}{dx^2} = -2 < 0$ , confirming a maximum.

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- Jamb Mathematics- Lesson notes on integration for utme Success

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Application of differentiation | Jamb Mathematics

Calculation problems involving rate of change

Problem 1: Rate of Change of Area of a Circle

Problem 2: Rate of Change of Volume of a Sphere

Problem 3: Rate of Change of Volume of a Cube

Problem 4: Ladder Sliding Down a Wall

Problem 5: Water Draining from a Conical Tank

Problem 6: Expanding Square

Problem 7: Expanding Rectangle

Problem 8: Shrinking Circular Pool

Problem 9: Draining Cylindrical Tank

Problem 10: Moving Particle

Problem 11: Expanding Cylinder

Problem 12: Melting Ice Sphere

Problem 13: Changing Angle of Elevation

Calculation problems involving maxima and minima in differentiation

Problem 1: Maximum Area of a Rectangular Enclosure

Problem 2: Minimum Surface Area of a Cylinder

Problem 3: Maximum Volume of a Box with Square Base

Problem 4: Minimum Distance from a Point to a Line

Problem 5: Maximum Profit

I recommend you check my Post on the following:

POSCHOLARS TEAM.

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