Find the approximate value of (4.01)3 Class 12 th sci msb
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I think question should be (4.01)½
we have to find approximate value of (4.01)½
consider, y = x½
differentiate with respect to x,
dy/dx = 1/2√x ......(1)
Let x = 4.00 , ∆x = 0.01
∆y = (x + ∆x)½ - x½
= (4.00 + 0.01)² - (4.00)½
= (4.01)½ - 2.00
∆y + 2.00. = (4.01)½
now , dy is approximately equals to ∆y and is given by
dy = (dy/dx)∆x
= (1/2√x)∆x [ from equation (1), ]
= {1/2√(4.00)} × 0.01
= (1/4) × 0.01
= 0.25 × 0.01
= 0.0025
hence, (4.01)½ = 2.00 + 0.0025
= 2.0025
we have to find approximate value of (4.01)½
consider, y = x½
differentiate with respect to x,
dy/dx = 1/2√x ......(1)
Let x = 4.00 , ∆x = 0.01
∆y = (x + ∆x)½ - x½
= (4.00 + 0.01)² - (4.00)½
= (4.01)½ - 2.00
∆y + 2.00. = (4.01)½
now , dy is approximately equals to ∆y and is given by
dy = (dy/dx)∆x
= (1/2√x)∆x [ from equation (1), ]
= {1/2√(4.00)} × 0.01
= (1/4) × 0.01
= 0.25 × 0.01
= 0.0025
hence, (4.01)½ = 2.00 + 0.0025
= 2.0025
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