Tutorial : Determining Approximate Values Problem 2

Tutorial — By Christopher on February 20, 2009 at 12:00 AM

Given that y = 4x² – 5x + 3. If x increases by 1.5% at the instant when x = 3, find the corresponding percentage change in y.

Taken from the Integrated Curriculum for Secondary Schools Additional Mathematics Form 4 2005 ed. pg 215 Formative Exercise 9.5 Question 8

Recall that y_new = y_initial + δy = y_initial + (dy/dx)δx.
Note that δy/δx = dy/dx.
Hence, δy = (dy/dx)δx.

First, we find dy/dx.
y = 4x² – 5x + 3
dy/dx = (2)4x^{2 – 1} – 5x^{1 – 1} = 8x – 5

δy = (dy/dx)δx.
Find the value of δx .
We know that x increased by 1.5% at the instant when x = 3.
Therefore,
δx = 3 x 0.015 = 0.045.

Find the value of dy/dx at the instant when x = 3
Therefore,
dy/dx = 8(3) – 5 = 19.

Replace dy/dx and δx with the acquired values.
δy = (19)(0.045) = 0.855

At the instant when x = 3,
y = 4(3)² – 5(3) + 3 = 24

Therefore, the percentage of change in y = 0.855 / 24 X 100% = 3.56%

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