0606 P12 - Mar 2022 - Q5 - 7 marks
7752
Variables \(x\) and \(y\) are such that \(\displaystyle y=\frac{\ln(2x^2-3)}{3x}\).
(a) Find \(\displaystyle \frac{dy}{dx}\).
(b) Hence find the approximate change in \(y\) when \(x\) increases from 2 to \(2+h\), where \(h\) is small.
(c) At the instant when \(x=2\), \(y\) is increasing at the rate of 4 units per second. Find the corresponding rate of increase in \(x\).
Solution
Answer: (a) \(\displaystyle \frac{4}{3(2x^2-3)}-\frac{\ln(2x^2-3)}{3x^2}\). (b) \(0.133h\). (c) \(30.2\) units per second.
Answer: (a) \(\displaystyle \frac{4}{3(2x^2-3)}-\frac{\ln(2x^2-3)}{3x^2}\). (b) \(0.133h\). (c) \(30.2\) units per second.
(a) Write
\(\displaystyle y=\frac{\ln(2x^2-3)}{3x}\).
Using the quotient rule,
\(\displaystyle \frac{dy}{dx}=\frac{3x\cdot\frac{4x}{2x^2-3}-3\ln(2x^2-3)}{9x^2}.\)
This simplifies to
\(\displaystyle \frac{dy}{dx}=\frac{4}{3(2x^2-3)}-\frac{\ln(2x^2-3)}{3x^2}.\)
(b) At \(x=2\),
\(\displaystyle \frac{dy}{dx}=\frac{4}{3(5)}-\frac{\ln5}{12}\approx0.133.\)
For a small increase \(h\) in \(x\), the approximate change in \(y\) is
\(\displaystyle \Delta y\approx\frac{dy}{dx}h\approx0.133h\).
(c) Since
\(\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt},\)
we have
\(\displaystyle 4\approx0.133\frac{dx}{dt}\).
Therefore
\(\displaystyle \frac{dx}{dt}\approx\frac{4}{0.133}=30.2\).
