In the previous section, we completed a discussion on differential approximation. In this section, we will see how it can be used to find amount of error.
The basic details can be demonstrated using an example. It can be written in 6 steps:
1. We know that, volume of a sphere can be obtained using the formula: $\rm{V = \frac{4}{3} \pi r^3}$
♦ V is the volume of the sphere.
♦ r is the radius of the sphere.
2. Based on the above formula, we can say that, volume is a function of radius. That is.,
$\rm{V = f(r) = \frac{4}{3} \pi r^3}$
3. Suppose that, the radius of a sphere is measured to be 5 cm with an error of 0.1 cm.
•
Then the actual radius r will be such that:
4.9 ≤ r ≤ 5.1
4. Consider the situation:
♦ The actual radius is 5.1 cm.
♦ We use 5 cm for calculating the volume.
•
In such a situation, there will be error in the calculated volume.
•
How much error will be present in the calculated volume?
Answer can be written in 4 steps:
(i) Calculated volume =
$\rm{V_1 \,=\, f(r_1) \,=\, \frac{4}{3} \pi {r_1}^3 \,=\, \frac{4}{3} \pi (5^3)}$
(ii) Actual volume =
$\rm{V_2 \,=\, f(r_2) \,=\, \frac{4}{3} \pi {r_2}^3 \,=\, \frac{4}{3} \pi ({5.1}^3)}$
(iii) So the error in the calculated volume =
$\rm{V_2 - V_1 \,=\, \Delta V \,=\, \frac{4}{3} \pi ({5.1}^3) ~-~\frac{4}{3} \pi ({5}^3)}$
(iv) But based on differential approximation that we learned in the previous section, we need not find the exact ΔV. we can write:
♦ ΔV ≈ dV (This is possible because, r2 is close to r1)
♦ dV = f'(r1).dr
•
Thus we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(r_1) . dr} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{4}{3} \pi (3 {r_1}^2) . (r_2 - r_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{4 \pi (5^2) . (5.1 - 5)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{4 \pi (25) . (0.1)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{10 \pi~\rm{cm^3}} \\
\end{array}$
5. Consider the situation:
♦ The actual radius is 4.9 cm.
♦ We use 5 cm for calculating the volume.
•
In such a situation also, there will be error in the calculated volume.
•
How much error will be present in the calculated volume?
Answer can be written in 4 steps:
(i) Calculated volume =
$\rm{V_1 \,=\, f(r_1) \,=\, \frac{4}{3} \pi {r_1}^3 \,=\, \frac{4}{3} \pi (5^3)}$
(ii) Actual volume =
$\rm{V_2 \,=\, f(r_2) \,=\, \frac{4}{3} \pi {r_2}^3 \,=\, \frac{4}{3} \pi ({4.9}^3)}$
(iii) So the error in the calculated volume =
$\rm{V_2 - V_1 \,=\, \Delta V \,=\, \frac{4}{3} \pi ({4.9}^3) ~-~\frac{4}{3} \pi ({5}^3)}$
(iv) But based on differential approximation that we learned in the previous section, we need not find the exact ΔV. we can write:
♦ ΔV ≈ dV (This is possible because, r2 is close to r1)
♦ dV = f'(r1).dr
•
Thus we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(r_1) . dr} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{4}{3} \pi (3 {r_1}^2) . (r_2 - r_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{4 \pi (5^2) . (4.9 - 5)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{4 \pi (25) . (-0.1)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{-10 \pi~\rm{cm^3}} \\
\end{array}$
6. So we can write:
The error (dV) in calculated volume will be such that:
−10π ≤ dV ≤ 10π.
Let us cross check by calculating ΔV
•
Consider the situation:
♦ The actual radius is 5.1 cm.
♦ We use 5 cm for calculating the volume.
Then we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\Delta V} & {~=~} &{V_2 - V_1} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{4}{3} \pi ({5.1}^3)~-~\frac{4}{3} \pi ({5}^3)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{10.2013333333333 \pi ~\rm{cm^3}} \\
\end{array}$
•
Consider the situation:
♦ The actual radius is 5.1 cm.
♦ We use 5 cm for calculating the volume.
Then we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\Delta V} & {~=~} &{V_2 - V_1} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{4}{3} \pi ({4.9}^3)~-~\frac{4}{3} \pi ({5}^3)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{-9.80133333333329 \pi ~\rm{cm^3}} \\
\end{array}$
•
When we use differential approximation, we get 10π and −10π.
•
10π is approximately equal to 10.2013 π.
• −10π is approximately equal to −9.8013333 π.
Solved example 22.37
If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.
Solution:
1. The actual radius r will be such that:
8.97 ≤ r ≤ 9.03
2. Consider the situation:
♦ The actual radius is 9.03 cm.
♦ We use 9 cm for calculating the volume.
3. First we fix r1 and r2:
r2 must be the value which causes the difficulty. So we put:
♦ r2 = 9.03 cm
♦ r1 = 9.0 cm
4. Then we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(r_1) . dr} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{4}{3} \pi (3 {r_1}^2) . (r_2 - r_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{4 \pi (9^2) . (9.03 - 9)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{4 \pi (81) . (0.03)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{9.72 \pi~\rm{cm^3}} \\
\end{array}$
5. Consider the situation:
♦ The actual radius is 8.97 cm.
♦ We use 9 cm for calculating the volume.
•
Here we get the same result with opposite sign because dx is the same value but with opposite sign.
•
That means, dV in this case is −9.72π cm3.
6. We can write:
The error (dV) in calculated volume will be such that:
−9.72π ≤ dV ≤ 9.72π.
Solved example 22.38
If the length of a cube is measured as 6
cm with an error of 0.2 cm, then find the approximate error in
calculating its volume.
Solution:
1. The actual length (l) will be such that:
5.8 ≤ l ≤ 6.2
2. Consider the situation:
♦ The actual length is 6.2 cm.
♦ We use 6 cm for calculating the volume.
3. First we fix l1 and l2:
l2 must be the value which causes the difficulty. So we put:
♦ l2 = 6.2 cm
♦ l1 = 6.0 cm
4. Then we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{dV} & {~=~} &{f'(l_1) . dl} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{ (3 {l_1}^2) . (l_2 - l_1)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{3 (6^2) . (6.2 - 6)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{3 (36) . (0.2)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{21.6~\rm{cm^3}} \\
\end{array}$
5. Consider the situation:
♦ The actual length is 5.8 cm.
♦ We use 6 cm for calculating the volume.
•
Here we get the same result with opposite sign because dx is the same value but with opposite sign.
•
That means, dV in this case is −21.6 cm3.
6. We can write:
The error (dV) in calculated volume will be such that:
−21.6 ≤ dV ≤ 21.6.
In the next section, we will see change in quantity.
Previous
Contents
Next
Copyright©2024 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment