These are the solutions to the WASSCE Further Mathematics past questions on Differential Calculus.
The TI-84 Plus CE shall be used for applicable questions.
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Difference Quotient and Derivative (Derivatives by Limits)
$
a, n \:\:are\:\:constants \\[3ex]
(1.)\:\: \underline{Power\;\;Rule} \\[3ex]
y = ax^n \\[3ex]
\dfrac{dy}{dx} = nax^{n - 1} \\[7ex]
(2.)\:\: \underline{Sum/Difference\;\;Rule} \\[3ex]
y = u \pm v \pm w \\[3ex]
u = f(x);\:\: v = f(x);\;\; w = f(x) \\[3ex]
\dfrac{dy}{dx} = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \pm \dfrac{dw}{dx} \\[7ex]
(3.)\:\: \underline{Chain\;\;Rule} \\[3ex]
y = f(u) \\[3ex]
u = f(x) \\[3ex]
\dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} \\[7ex]
Also: \\[3ex]
y = f(u) \\[3ex]
u = f(v) \\[3ex]
v = f(x) \\[3ex]
\dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dx} \\[7ex]
Also: \\[3ex]
y = f(u) \\[3ex]
u = f(v) \\[3ex]
v = f(w) \\[3ex]
w = f(x) \\[3ex]
\dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dw} * \dfrac{dw}{dx} \\[7ex]
...and\;\;so\;\;on\;\;and\;\;so\;\;forth \\[5ex]
(4.)\:\: \underline{Product\;\;Rule} \\[3ex]
y = u * v \\[3ex]
u = f(x);\:\: v = f(x) \\[3ex]
\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \\[7ex]
(5.)\:\: \underline{Quotient\;\;Rule} \\[3ex]
y = \dfrac{u}{v} \\[5ex]
u = f(x);\:\: v = f(x) \\[3ex]
\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}
$
Standard Derivatives of Exponential Functions
$
a\;\; is\:\:a\;\;positive\:\:constant \\[3ex]
k \:\:is\:\:a\;\:constant \\[3ex]
(1.)\:\: y = e^x \\[3ex]
\dfrac{dy}{dx} = e^x \\[7ex]
(2.)\;\; y = e^{kx} \\[3ex]
\dfrac{dy}{dx} = ke^{kx} \\[7ex]
(3.)\;\; y = e^{-kx} \\[3ex]
\dfrac{dy}{dx} = -ke^{kx} \\[7ex]
(4.)\:\: y = a^x \\[3ex]
\dfrac{dy}{dx} = a^x \ln a \\[7ex]
$
Standard Derivatives of Logarithmic Functions
$
a\;\; is\:\:a\;\;positive\:\:constant \\[3ex]
k \:\:is\:\:a\;\:constant \\[3ex]
(1.)\:\: y = \ln x \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{x} \\[7ex]
(2.)\:\: y = \log_a x \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{x \ln a} \\[7ex]
(3.)\:\: y = \ln |x| \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{x} \\[7ex]
(4.)\:\: y = \log_a |x| \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{x \ln a} \\[7ex]
$
Standard Derivatives of Trigonometric Functions
$
(1.)\:\: y = \sin x \\[3ex]
\dfrac{dy}{dx} = \cos x \\[7ex]
(2.)\:\: y = \cos x \\[3ex]
\dfrac{dy}{dx} = -\sin x \\[7ex]
(3.)\:\: y = \tan x \\[3ex]
\dfrac{dy}{dx} = \sec^2 x \\[7ex]
(4.)\:\: y = \csc x \\[3ex]
\dfrac{dy}{dx} = -\csc x \cot x \\[7ex]
(5.)\:\: y = \sec x \\[3ex]
\dfrac{dy}{dx} = \sec x \tan x \\[7ex]
(6.)\:\: y = \cot x \\[3ex]
\dfrac{dy}{dx} = -\csc^2 x
$
Standard Derivatives of Inverse Trigonometric Functions
$
(1.)\:\: y = \sin^{-1} x \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}} \\[7ex]
(2.)\:\: y = \cos^{-1} x \\[3ex]
\dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1 - x^2}} \\[7ex]
(3.)\;\; y = \tan^{-1} x \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{1 + x^2} \\[7ex]
(4.)\:\: y = \csc^{-1} x \\[3ex]
\dfrac{dy}{dx} = \dfrac{-1}{|x|\sqrt{x^2 - 1}} \\[7ex]
(5.)\:\: y = \sec^{-1} x \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{|x|\sqrt{x^2 - 1}} \\[7ex]
(6.)\;\; y = \cot^{-1} x \\[3ex]
\dfrac{dy}{dx} = \dfrac{-1}{1 + x^2} \\[7ex]
$
$
(a.) \\[3ex]
MC = TC'(x) \\[3ex]
where \\[3ex]
x = \text{number of items} \\[3ex]
MC = \text{marginal cost function} \\[3ex]
TC = \text{total cost function} \\[5ex]
(b.) \\[3ex]
MR = TR'(x) \\[3ex]
where \\[3ex]
x = \text{number of items} \\[3ex]
MR = \text{marginal revenue function} \\[3ex]
TR = \text{total revenue function} \\[5ex]
(c.) \\[3ex]
MPC = NC'(x) \\[3ex]
where \\[3ex]
x = \text{disposable national income} \\[3ex]
MPC = \text{marginal propensity to consume} \\[3ex]
NC = \text{national consumption function} \\[5ex]
$
(2.)
(1.) A curve is given by $y = 8x + \dfrac{27}{2x^2}$.
Find:
(a.) an expression for $\dfrac{dy}{dx}$;
(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)
$
(a.) \\[3ex]
y = 8x + \dfrac{27}{2x^2} \\[5ex]
= 8x + \dfrac{27}{2} * \dfrac{1}{x^2} \\[5ex]
= 8x^1 + \dfrac{27}{2} * x^{-2} \\[5ex]
\dfrac{dy}{dx} = 1 * 8x^{1 - 1} + -2 * \dfrac{27}{2} * x^{-2 - 1} \\[5ex]
= 8x^0 + -27x^{-3} \\[3ex]
= 8(1) - \dfrac{27}{x^3} \\[5ex]
= 8 - \dfrac{27}{x^3} \\[5ex]
(b.) \\[3ex]
\dfrac{dy}{dx} = 0 \\[5ex]
8 - \dfrac{27}{x^3} = 0 \\[5ex]
8 = \dfrac{27}{x^3} \\[5ex]
8x^3 = 27 \\[3ex]
x^3 = \dfrac{27}{8} \\[5ex]
x = \sqrt[3]{\dfrac{27}{8}} \\[5ex]
x = \dfrac{3}{2} \\[5ex]
y = 8x + \dfrac{27}{2x^2} \\[5ex]
y = 8x + 27 \div 2x^2 \\[5ex]
@\; x = \dfrac{3}{2} \\[5ex]
y = 8\left(\dfrac{3}{2}\right) + 27 \div 2\left(\dfrac{3}{2}\right)^2 \\[5ex]
= 4(3) + 27 \div \dfrac{3^2}{2} \\[5ex]
= 12 + 27 \div \dfrac{9}{2} \\[5ex]
= 12 + 27 * \dfrac{2}{9} \\[5ex]
= 12 + 6 \\[3ex]
= 18 \\[3ex]
\text{Stationary Point} = (x, y) = \left(\dfrac{3}{2}, 18\right) \\[5ex]
$
To determine the nature of the stationary point, let us use the Second Derivative Test
If the second derivative is positive at the stationary point, the graph of the function is concave upward,
and the point is a local minimum.
If the second derivative is negative, the graph of the function is concave downward, and the point is a
local maximum.
$
\underline{\text{Nature of the Stationary Point: 2nd Derivative Test}} \\[3ex]
\dfrac{dy}{dx} = 8 - \dfrac{27}{x^3} \\[5ex]
\dfrac{dy}{dx} = 8 - 27x^{-3} \\[5ex]
\dfrac{d^2y}{dx^2} = -3 * -27 * x^{-3 - 1} \\[5ex]
= 81 * x^{-4} \\[3ex]
= \dfrac{81}{x^4} \\[5ex]
= 81 \div x^4 \\[3ex]
@\; x = \dfrac{3}{2} \\[5ex]
\dfrac{d^2y}{dx^2} = 81 \div \left(\dfrac{3}{2}\right)^4 \\[5ex]
= 81 \div \dfrac{3^4}{2^4} \\[5ex]
= 81 \div \dfrac{81}{16} \\[5ex]
= 81 * \dfrac{16}{81} \\[5ex]
= 16 \\[3ex]
16 \gt 0 \\[3ex]
\therefore \left(\dfrac{3}{2}, 18\right) \text{is a minimum point} \\[5ex]
$
Let: m1 = slope of the tangent to the curve = derivative m2 = slope of the normal to the curve
The product of the two slopes is −1 because the normal to the curve is perpendicular to the tangent to
the curve
(2.) Find the equation of the normal to the curve $y = 7x - 5x^2$ at x = 2.
1st: We shall find the slope of the tangent to the curve at at x = 2.
This is the derivative of the function at x = 2.
2nd: We shall determine the slope of the normal to the curve
The normal to the curve is perpendicular to the tangent to the curve
Hence, the product of the two slopes (product of the slope of the tangent and the slope of the normal) is
−1
3rd: We shall determine the point (x1, y1) on the curve
We already know the x-coordinate, but we shall find the y-coordinate
4th: We shall use the Point-Slope Form to determine the equation of the normal to the curve.
$
y = 7x - 5x^2 \\[3ex]
\underline{\text{Slope of the tangent to the curve}}, m_T \\[3ex]
\dfrac{dy}{dx} = 7 - 10x \\[3ex]
@\;x = 2 \\[3ex]
m_T = \left.\dfrac{dy}{dx}\right|_{x = 2} \\[5ex]
= 7 - 10(2) \\[3ex]
= 7 - 20 \\[3ex]
= -13 \\[5ex]
\underline{\text{Slope of the normal to the curve}}, m_N \\[3ex]
m_N * m_T = -1 \\[3ex]
m_N * 13 = -1 \\[3ex]
m_N = \dfrac{-1}{-13} \\[5ex]
m_N = \dfrac{1}{13} \\[5ex]
\underline{\text{Point on the curve}}, (x_1, y_1) \\[3ex]
y = 7x - 5x^2 \\[3ex]
@\;x = 2 \\[3ex]
y = 7(2) - 5(2)^2 \\[3ex]
= 14 - 5(4) \\[3ex]
= 14 - 20 \\[3ex]
= -6 \\[3ex]
(x_1, y_1) = (2, -6) \\[5ex]
\underline{\text{Equation of the normal to the curve}} \\[3ex]
\text{Point – Slope Form} \\[3ex]
y - y_1 = m_N(x - x_1) \\[4ex]
y - -6 = \dfrac{1}{13}(x - 2) \\[5ex]
y + 6 = \dfrac{1}{13}x - \dfrac{2}{13} \\[5ex]
y = \dfrac{1}{13}x - \dfrac{2}{13} - 6 \\[5ex]
y = \dfrac{1}{13}x - \dfrac{2}{13} - \dfrac{78}{13} \\[5ex]
y = \dfrac{1}{13}x - \dfrac{80}{13} \\[5ex]
LCD = 13 \\[3ex]
\text{Multiply each term by 13} \\[3ex]
13y = 13\left(\dfrac{1}{13}x\right) - 13\left(\dfrac{80}{13}\right) \\[5ex]
13y = x - 80 \\[3ex]
13y - x + 80 = 0
$
(3.)
(4.) A boy runs in a line and his displacement at time t seconds after leaving the start point O is X
metres, where $20X = 4t^2 + t^3$.
Find the:
(i.) velocity of the boy when t = 15 seconds;
(ii.) value of t for which the acceleration of the boy is 8 times his initial acceleration.
