1. If y = x2 + 3x + 6, x = 10, δx = 0.01 then find y and df
Sol: Given y = x2 + 3x + 6
x = 10
δx = 0.01
⇒ δy = f(10 + 0.01) - f(10)
⇒ δy = f(10.01) - f(10)
⇒ δy = [(10.01)2 + 3 (10.01) + 6] - [(10)2 + 3(10) + 6]
⇒ δy = 100.2001 + 30.03 + 6 - 100 - 30 - 6
⇒ δy = 0.2001 + 0.03
∴ δy = 0.2301
⇒ dy = (2x + 3) x
⇒ dy = [2(10) + 3] (0.01)
∴ dy = 0.23
2. If y = log x, x = 3, δx = 0.003 then find y and dy
Sol. Given y = log x
x = 3
δx = 0.003
⇒ δy = f (3 + 0.003) - f (3)
δy = f (3.003) - f (3)
δy = log (3.003) - log 3
δy = log (
)
∴ δy = log (1.001)
3. Find the approximate value of 
Sol: Let: f(x + δx) =
δx = 81
δx = 1
f(x) = 
4. Find the approximate value of 
Sol: Let: f(x + δx) =
x = 64
δx = - 1
f(x) = 
5. Find the approximate value of 
Sol: Let : f(x +δx) =
x = 25
δx = 0.2
f(x) =
6. Find the approximate value of 
sol: Let: f (x + δx) =
x = 8 ; δ x = -0.2
f(x) =
7. Find the approximate value of sin 60º 1' (Given that 1' = 0.1745 radian).
sol. Let: f (x + δx) = sin 60º 1'
x = 60º
f(x) = sin x
8. Find the approximate value of cos 45º 6'. (Given that 1' = 0.1745 radian).
Sol. Let: f (x + δx) = cos 45º 6'
x = 45º
f(x) = cosx
9. If the increase in the side of a square is 1% then find the percentage of change in the area of 
the square.
Sol: Let 'x' be the side and 'A' be the area of square.
Area of square: A = x2
10. The diameter of a sphere is measured to be 20 cm. If an error of 0.02 cm occurs in this, find the errors in volume and surface area of the sphere.
Sol: Let 'd' be the diameter, 'V' be the volume and 'S' be the surface area of sphere.
Given: d = 20 ; δd = 0.02 ; δV, δS = ?
Volume of sphere:
∴ Error in volume : δV : 4 πcm3
Surface area of sphere: S = πd2
⇒ δS = 2πd . δd
⇒ δS = 2π (20) (0.02)
∴ Error in surface area: δS = 0.8π sq.cm.
11. Area of a ∆ABC is measured by the measures of a, b, c. If c is the error in measuring c, then what is the percentage error in the area.
sol: Let 's' be the area of triangle.
Given: S = 
ab sinC 
⇒ δs = ab cos C δc
Percentage error in Area:
= 100 cotC . δc
12. The time of a complete oscillation of a simple pendulum of length l is given by the equation 
where g is gravitational constant. Find the approximate error in the calculated g corresponding to an error of 0.01 percent in the value of t.
Sol: Given:
Given :
Taking " log" on both sides
∴ Percentage error in g : - 0.02.
Writer Sayyad Anwar
