Modules › Optimization, L'Hôpital's rule, Newton's method, antiderivatives
Optimization, L'Hôpital's rule, Newton's method, antiderivatives
Applied optimization, indeterminate forms, Newton's iteration and antiderivatives.
Lecture 6 · Tutorial 6 · Applications of derivatives and antiderivatives
Textbook sections: Thomas ch. 4 Applications of Derivatives (4.1–4.8) · Stewart ch. 4 Applications of Differentiation (4.1–4.9)
Textbook sections: Thomas ch. 4 Applications of Derivatives (4.1–4.8) · Stewart ch. 4 Applications of Differentiation (4.1–4.9)
Optimization, L'Hôpital's rule, Newton's method, antiderivatives
Calculus I · Module 7
After this module you can
- Solve an optimization problem in six steps and justify the maximum or minimum.
- Resolve $\frac00$, $\frac\infty\infty$, $0\cdot\infty$, $\infty-\infty$, $1^\infty$, $0^0$, $\infty^0$ with L'Hôpital.
- Run Newton's method by hand; find antiderivatives and check by differentiation.
Optimization: six steps
- 1. Draw and name the variables. 2. Write the quantity $Q$ to optimise.
- 3. Use constraints to reduce $Q$ to one variable. 4. State the domain.
- 5. Solve $Q'=0$; compare with endpoints / use a derivative test.
- 6. Answer the question asked (dimensions, value, units). Example: open box from $8\times15$: $V=x(8-2x)(15-2x)$, $x=\frac53$, $V=\frac{2450}{27}$.
Classic results
| Problem | Answer |
|---|---|
| rectangle of area $16$, least perimeter | square $4\times4$, perimeter $16$ |
| fence $800$ m along a river | $200\times400$ m, area $80\,000$ |
| tank $500$ ft³, square base, open top, least steel | base $10$, height $5$ |
| rectangle under $y=27-x^2$ | $x=3$, area $108$ |
| rectangle in right isosceles triangle (hyp. $2$) | area $\frac12$ |
L'Hôpital's rule
- If $\frac f g$ is $\frac00$ or $\frac{\pm\infty}{\pm\infty}$ at $a$ and $\lim\frac{f'}{g'}$ exists, then $\lim\frac fg=\lim\frac{f'}{g'}$.
- $0\cdot\infty$: rewrite as a quotient: $x\ln x=\frac{\ln x}{1/x}\to0$. $\infty-\infty$: combine fractions.
- $1^\infty,0^0,\infty^0$: take $\ln$: $\left(1+\frac1x\right)^x\to e$, $x^{1/x}\to1$.
- $\lim_{x\to0}\frac{9x-3\sin3x}{5x^3}=\frac{27}{10}$ (three applications).
Newton's method
- Solve $f(x)=0$: $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ (follow the tangent to the $x$-axis).
- $f=x^2+x-1$, $x_0=-1$: $x_1=-2$, $x_2=-\frac53$; $x_0=1$: $x_1=\frac23$, $x_2=\frac{13}{21}$.
- Fails if $f'(x_n)=0$, or oscillates: $f=\sqrt{|x|}\operatorname{sgn}x$ gives $h\to-h\to h$.
- Choose $x_0$ from a sketch or a sign change.
Antiderivatives
- $F'=f$; all antiderivatives are $F+C$. Check by differentiating.
- $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$ ($n\ne-1$), $\int\frac{dx}{x}=\ln|x|+C$, $\int\cos x=\sin x$, $\int\sin x=-\cos x$.
- $\int\csc x\cot x\,dx=-\csc x+C$; $\int\sec^2x\,dx=\tan x+C$.
- Rewrite first: $\frac{t\sqrt t+\sqrt t}{t^2}=t^{-1/2}+t^{-3/2}$, antiderivative $2\sqrt t-\frac{2}{\sqrt t}$.
Typical mistakes
OptimizationForgetting the domain or the second check (min or max?).
L'HôpitalUsing it when the form is not indeterminate (e.g. $\frac{1}{0^+}$).
AntiderivativesDropping $+C$ or the factor $\frac1a$ for $\cos(ax)$.
Practice next
- Chapter quizzes Ch. 3 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Newton's method and bisection
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Optimization (Tutorial 6, Q2)
A rectangle with base on the hypotenuse (length $2$) is inscribed in an isosceles right triangle. The largest possible area is
Show the answer and the solution
Answer: $\frac{1}{2}$
Put the hypotenuse on $[-1,1]$ with apex $(0,1)$: a rectangle of half-width $x$ has height $1-x$, so $A=2x(1-x)$, maximal at $x=\frac12$: $A=\frac12$ (dimensions $1\times\frac12$).
Worked example 2 · L'Hôpital's rule (Tutorial 6, Q7)
For which $c$ is $f(x)=\dfrac{9x-3\sin3x}{5x^3}$ ($x\ne0$), $f(0)=c$, continuous at $0$?
Show the answer and the solution
Answer: $\frac{27}{10}$
Need $c=\lim_{x\to0}f$: L'Hôpital three times, $\frac{81\cos3x\cdot}{\cdots}$ — equivalently $3\sin3x\approx9x-\frac{27}2x^3$, so the numerator $\approx\frac{27}{2}x^3$ and the limit is $\frac{27/2}{5}=\frac{27}{10}$.
Worked example 3 · Newton's method (Tutorial 6, Q10)
For $f(x)=\sqrt{|x|}\operatorname{sgn}(x)$ (i.e. $\sqrt x$ for $x\ge0$, $-\sqrt{-x}$ for $x<0$), Newton's method from $x_0=h>0$ gives $x_1=$
Show the answer and the solution
Answer: $-h$ (and then $x_2=h$: the iteration oscillates)
For $x>0$: $x_1=h-\frac{\sqrt h}{1/(2\sqrt h)}=h-2h=-h$; by symmetry the next step returns to $h$.
Worked example 4 · Antiderivatives (Tutorial 6, Q11(a))
Find the most general antiderivative of $\dfrac{t\sqrt t+\sqrt t}{t^2}$.
Show the answer and the solution
Answer: $C + 2 \sqrt{t} - \frac{2}{\sqrt{t}}$
Split: $\frac{t^{3/2}+t^{1/2}}{t^2}=t^{-1/2}+t^{-3/2}$, so $F=2t^{1/2}-2t^{-1/2}+C$. Check by differentiating.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Newton's method and bisection