1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.

(xiv) ( 3.968) ^{3/2 }
 

Answers (1)

Let's suppose y = (x)^{\frac{3}{2}} and let x = 4 and \Delta x = -0.032
Then,
\Delta y = ({x+\Delta x})^{\frac{3}{2}} - (x)^{\frac{3}{2}}
\Delta y = ({4 - 0.032})^{\frac{3}{2}} - (4)^{\frac{3}{2}}
\Delta y = ({3.968})^{\frac{3}{2}} - 8
({3.968})^{\frac{3}{2}} = \Delta y + 8
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{3 (x)^{\frac{1}{2}}}{2}.(-0.032) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{3}{2}} \ and \ \Delta x = -0.032)\\ dy = \frac{3 (4)^{\frac{1}{2}}}{2}.(-0.032)\\ dy = \frac{3\times 2}{2}.(-0.032)\\\\dy = -0.096
Now,
(3.968)^{\frac{3}{2}} = \Delta y +8\\ (3.968)^{\frac{3}{2}} = (-0.096) + 8\\ (3.968)^{\frac{3}{2}} = 7.904
Hence, (3.968)^{\frac{3}{2}} is approximately equal to 7.904  


        

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Foundation 2021 Class 9th Maths

Master Maths with "Foundation course for class 9th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions