SECTION B (PART 1)
A curve is given by y = 8x + \(\frac{27}{2x^2}\),
FIND:
(a) an expression for \(\frac{dy}{dx}\),
(b) the coordinates of the stationary point on the curve and the nature of the stationary point;
(c) the equation of the normal to the curve at (2, 2).
Explanation
Given: y = 8x + \(\frac{27}{2x^2}\) = y = 8x + \(\frac{27}{2}\)x\(^{-2}\),
(a) \(\frac{dy}{dx}\) = 8x\(^0\) + \(\frac{27}{2}\)(-2)x\(^{-3}\)
= 8 - 27x\(^{-3}\) = 8 - \(\frac{27}{x^3}\)
(b) \(\frac{dy}{dx}\) = 8 - \(\frac{27}{x^3}\) = 0
= 8 = \(\frac{27}{x^3}\)
= 8x\(^3\) = 27
x\(^3\) = \(\frac{27}{8}\)
x = \( \sqrt[3]{\frac{27}{8}}\) = \(\frac{3}{2}\)
x = \(\frac{3}{2}\)
y = 8x + \(\frac{27}{2x^2}\) = 8(\(\frac{3}{2}\)) + \(\frac{27}{2}\)(\(\frac{3}{2}\))\(^{-2}\)
y = 4(3) + \(\frac{27}{2}\)(\(\frac{4}{9}\))
y = 12 + 6 = 18
Thus, stationary points are (\(\frac{3}{2}\), 18)
\(\frac{d^2y}{dx^2}\) = - \(\frac{27}{x^3}\) = - 27(-3)x\(^{-4}\) = \(\frac{81}{x^4}\) = \(\frac{81}{(\frac{3}{2})^4}\) = 81 x \(\frac{16}{81}\) = 16 > 0
For all x, the stationary point is a minimum.
(c) gradient = 8 - \(\frac{27}{x^3}\) = 8 - \(\frac{27}{2^3}\) = 8 - \(\frac{27}{8}\) = \(\frac{37}{8}\)
Equation of normal, at (2, 2)
but m\(_2\) = \(\frac{- 1}{m_1}\) = \(\frac{- 1}{\frac{37}{8}}\) = \(\frac{- 8}{37}\)
y - y\(_1\) = m\(_2\)[x - x\(_1\)]
y - 2 = \(\frac{- 8}{37}\)[x - 2]
37y - 74 = - 8x + 16
37y + 8x - 90 = 0