Application of Derivatives Exercise 16.1 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to study

1. Application of Derivatives: Differentials

Exercise 16.1

1. Compute $\Delta $y, dy and $\Delta $y-dy when y = $\frac{{{{\rm{x}}^2}}}{2}$ + 3x. x = 2and dx = 0.5

Solution:

x = 2

Or, $\Delta $x = dx = 0.5

Or, y = $\frac{{{{\rm{x}}^2}}}{2}$ + 3x.

Or, y + $\Delta $y = $\frac{1}{2}$(x + $\Delta $x)² + 3(x + $\Delta $x).

Or, $\Delta $y = $\frac{1}{2}$(x + $\Delta $x)² + 3(x + $\Delta $x) – ($\frac{{{{\rm{x}}^2}}}{2}$ + 3x).

= x.${\rm{\: }}\Delta $x + $\frac{1}{2}{\left( {\Delta {\rm{x}}} \right)^2}$ + 3$\Delta $x = 2 * 0.5 + $\frac{1}{2}$ * (0.5)² + 3 * 0.5

= 1.0 + 0.125 + 1.5 = 2.625.

y = f(x) = $\frac{{{{\rm{x}}^2}}}{2}$ + 3x.

f’(x) = x + 3

dy = f’(x).dx = (x + 3).dx = (2 + 3) * 0.5 = 2.5

or, $\Delta $y – dy = 2.625 – 2.5 = 0.125.

2. Find an approximate change in the volume of a cube of side x m, caused by increasing the sides by 1%. What is the percentage increment in the volume?

Solution:

v = Volume of cube = x³.

$\Delta $x = dx = 1% of x = 0.01x.

Approximate increase in volume,

dv = 3x².dx = 3 * x² * 0.01x = 0.03x³m².

% increase in volume = $\frac{{0.03{{\rm{x}}^3}}}{{{{\rm{x}}^3}}}$ * 100 = 3%. 

3. Use differentials to approximate the change in $x^3$ changes from 5 to 5.01.

Solution:

x = 5, $\Delta $x = dx = 5.01 – 5 = 0.01

y = x³

dy = 3x².dx = 3 * (5)² * 0.01 = 0.75.

 

4. Find an approximate change in 1/x as x changes from 1 to 0:98

Solution:

x = 1

Or, $\Delta $x = dx = 0.98 – 1 = - 0.02.

Or, y = $\frac{1}{{\rm{x}}}$.

Or, dy = $ - \frac{1}{{{{\rm{x}}^2}}}$.dx = $ - \frac{1}{{{{\left( 1 \right)}^2}}}$ * (-0.02) = 0.02. 

5. A circular copper plate is heated so that its radius increases from 5 cm to 5.06 cm. Find the approximate increase in area and also the actual increase in area.

Solution:

r = 5cms,

$\Delta $r = dr = 5.06 – 5 = 0.06.

A = πr².

dA = 2πr.dr = 2π * 5 * 0.06 = 0.6πcm².

So, approximate increase in area = 0.6π cm².

Again, A = πr².

A + $\Delta $A = π(r + $\Delta $r)²

Or, $\Delta $A = π(r² + 2r.${\rm{\: }}\Delta $r + ($\Delta $r)²) – πr² = π.${\rm{\: }}\Delta $r(2r + $\Delta $r)

= π * 0.06(2 * 5 + 0.06) = π * 0.06 * 10.06 = 0.6036 π cm².

So, actual increase in area = 0.6036 π cm². 

6. Find the approximate increase in the surface area of a cube if the edge increases from 10 to 10.01 cm. Also calculate the percentage error in the use of differential approximation.

Solution:

x = side of a cube = 10cms

$\Delta $x = dx = 10.01 – 10 = 0.01 cms

A = surface area = 6x².

dA = 12x.dx = 12 * 10 * 0.01 = 1.2cm².

So approximate increase in area = 1.2 cm².

Again,

A = 6x²

A + $\Delta $A = 6(x + $\Delta $x)².

$\Delta $A = 6[x² + 2x$\Delta $x + ($\Delta $x)²] – 6x² = 6$\Delta $x(2x + $\Delta $x)

= 6 * 0.01 (2 * 10 + 0.01) = 6 * 0.01 * 20.01 = 1.2006

Actual increase in area = 1.2006

Error = 1.2006 – 1.2 = 0.0006

% Error = $\frac{{0.0006}}{{600}}$ * 100 = 0.0001%. 

7. Find the approximate increase in the volume of a sphere when its radius increases from 2 to 2.1. Find also the actual increase and compare the two values.

Solution:

r = radius = 2; r + $\Delta $r = 2.1

or, $\Delta $r = 2.1 – r = 2.1 – 2 = 0.01.

v = $\frac{4}{3}$πr³.

dv = $\frac{4}{3}$.π.3r².dr = 4π * (2)² * 0.1 = 1.6π

So, approximate increase in volume = 1.6π.

v = $\frac{4}{3}$.πr³.

or, v + $\Delta $v = $\frac{4}{3}$.π(r + $\Delta $r)³.

So, $\Delta $v = $\frac{4}{3}$.π(r + $\Delta $r)³ – $\frac{4}{3}$.πr³ = $\frac{4}{3}$π[(r + $\Delta $r)³ – r³]

= $\frac{4}{3}$.π[(2.1)³ – (2)³] = $\frac{4}{3}$.π * (9.261 – 8).

Actual increase in volume = $\frac{4}{3}$π * 1.261 = $\frac{{5.044}}{3}$π.

Or, $\frac{{{\rm{dv}}}}{{\Delta {\rm{v}}}}$ = $\frac{{1.6{\rm{\pi }}}}{{5.004\frac{{\rm{\pi }}}{3}}}$ = 0.9516.

So, dv :$\Delta $v = 0.9516 : 1.