Mathematics I · Unit 4 · 4 hrs
Applications of Derivatives
Exam-focused notes for Applications of Derivatives (Mathematics I, MTH117): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- Curve sketching
- Review of maxima and minima of one variable
- Optimization problems
- Newton's method
Curve sketching
Find where the function $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ is increasing and where it is decreasing. [5]
- $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ $$f'(x) = 12x^3 - 12x^2 - 24x$$ $$12x^3 - 12x^2 - 24x = 0$$ $$12x(x^2 - x - 2) = 0$$ $$12x(x-2)(x+1) = 0$$ $$x = -1,\ 0,\ 2$$ Interval Test $12x$ $(x-2)$ $(x+1)$ $f'(x)$ Behaviour --------------------- $(-\infty,-1)$ $x=-2...
Full solved answer →Sketch the curve $y = x^2 + 1$ with the guidelines of sketching. If $z = xy^2 + y^3$, $x = \sin t$, $y = \cos t$, find $\frac{dz}{dt}$ at $t = 0$. [10+0]
Part 1: Curve $y = x^2 + 1$ Part 2: - $z = xy^2 + y^3$ - $x = \sin t$, $y = \cos t$ - Evaluate $\dfrac{dz}{dt}$ at $t = 0$ --- 1. Domain: Polynomial, so domain is $(-\infty, \infty)$. 2. Intercepts: - $y$-intercept: $x=0 \Rightarrow y = 1$, point $(0,1)$. -...
Full solved answer →Sketch the curve $y = x^3 + x$. [5]
As a polynomial, the domain is all real numbers $(-\infty, \infty)$. Since it is strictly increasing (shown below) and unbounded, the range is also $(-\infty, \infty)$. x-intercept (set $y=0$): $$x^3 + x = 0 \implies x(x^2+1) = 0$$ Since $x^2 + 1 0$, the on...
Full solved answer →Newton's method
Starting with $x_1 = 1$, find the third approximate $x_3$ to the root of the equation $x^3 - x - 5 = 0$. [5]
- Equation: $f(x) = x^3 - x - 5 = 0$ - Starting value: $x1 = 1$ - Required: $x3$ (third approximation) - Method implied: Newton-Raphson Iteration formula: $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 1$$ At $x1 = 1$: $$f(1) = 1 - 1 - 5 = -5$$...
Full solved answer →Use Newton's method to find $\sqrt[6]{2}$, correct to five decimal places. [5]
- Target: $\sqrt[6]{2} = 2^{1/6}$ - Required accuracy: 5 decimal places Let $x = 2^{1/6}$, so $x^6 = 2$, giving: $$f(x) = x^6 - 2 = 0, \qquad f'(x) = 6x^5$$ $$x{n+1} = xn - \frac{xn^6 - 2}{6xn^5} = \frac{5xn^6 + 2}{6xn^5}$$ Since $1^6 = 1 < 2 < 2^6 = 64$, a...
Full solved answer →Find the third approximation $x_3$ to the root of the equation $f(x) = x^3 - 2x - 7$, setting $x_1 = 2$. [5]
- $f(x) = x^3 - 2x - 7$ - Initial guess: $x1 = 2$ - Required: third approximation $x3$ $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$$ --- $$f(2) = 8 - 4 - 7 = -3$$ $$f'(2) = 3(4) - 2 = 10$$ $$x2 = 2 - \frac{-3}{10} = 2 + 0.3 = 2.3$$ --- $$f...
Full solved answer →Starting with $x_1 = 2$, find the third approximation $x_3$ to the root of the equation $x^3 - 2x - 5 = 0$. [5]
- Equation: $f(x) = x^3 - 2x - 5$ - Initial approximation: $x1 = 2$ - Required: third approximation $x3$ Iteration formula: $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$$ $$f(2) = 8 - 4 - 5 = -1$$ $$f'(2) = 3(4) - 2 = 10$$ $$x2 = 2 - \frac{...
Full solved answer →Optimization problems
A farmer has 2000 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area? Sketch the curve $y = \frac{1}{x-3}$ [5+5]
(a) Farmer's Fencing Optimization - Total fencing available: $2000$ ft - Shape: rectangular field bordering a straight river - No fence needed along the river side Let: - $x$ = each of the two sides perpendicular to the river (width) - $y$ = the side parall...
Full solved answer →Make Unit 4 stick
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