Given $y =x^3$, find the approximate increase in y when x increases from 2 to 2.02
(a) 0.24
(b) 0.12
(c) 0.48
(d) 0.04
Answer (1)
Find the derivative of the function y = x 3 , which is d x d y = 3 x 2 .
Evaluate the derivative at x = 2 : d x d y ∣ x = 2 = 3 ( 2 ) 2 = 12 .
Calculate the change in x : Δ x = 2.02 − 2 = 0.02 .
Approximate the change in y using the formula Δ y ≈ d x d y Δ x , which gives Δ y ≈ 12 × 0.02 = 0.24 .
Explanation
Problem Analysis We are given the function y = x 3 and asked to find the approximate increase in y when x increases from 2 to 2.02. This is a problem of finding the approximate change in a function using differentials.
Differential Approximation We can use the differential approximation formula: Δ y ≈ d x d y Δ x , where Δ y is the approximate change in y , d x d y is the derivative of y with respect to x , and Δ x is the change in x .
Finding the Derivative First, we find the derivative of y with respect to x :
d x d y = d x d ( x 3 ) = 3 x 2
Evaluating the Derivative Next, we evaluate the derivative at x = 2 :
d x d y ∣ x = 2 = 3 ( 2 ) 2 = 3 ( 4 ) = 12
Finding the Change in x We are given that x increases from 2 to 2.02, so the change in x is: Δ x = 2.02 − 2 = 0.02
Calculating the Approximate Change in y Now, we can calculate the approximate change in y :
Δ y ≈ d x d y Δ x = 12 × 0.02 = 0.24
Final Answer Therefore, the approximate increase in y when x increases from 2 to 2.02 is 0.24.
Examples
In physics, if you have a formula for the distance an object falls over time, like d = 5 t 2 , you can use this method to quickly estimate how much further the object falls in a tiny increment of time. For example, if you want to know how much further the object falls between 2 seconds and 2.01 seconds, you can use differentials to approximate the change in distance without having to calculate d at both times and subtract. This is useful when you need a quick estimate and don't need the exact value.
