Foundations of Mathematics: free practice, theory and problems

1. Numbers, fractions and percent

What is it about?

All engineering mathematics rests on ordinary arithmetic. Before you can differentiate, solve equations or calculate forces and currents, you must be able to work confidently with negative numbers, fractions, decimals and percentages – without guessing the order of operations. This unit starts from scratch and gives you the rules you will use in every other course.

Fractions and percentages are two ways of describing a part of a whole. Engineers use them all the time: efficiency, the gradient of a road, mixing ratios in concrete, tolerances and price changes.

Concepts and rules

• Order of operations: 1) parentheses, 2) powers, 3) multiplication and division (left to right), 4) addition and subtraction (left to right). Example: $2 + 3\cdot 4 = 2 + 12 = 14$.
• Negative numbers: minus times minus gives plus, and minus times plus gives minus. Subtracting a negative number is the same as adding: $5 - (-3) = 5 + 3 = 8$.
• Fraction: $\frac{a}{b}$ means $a$ divided by $b$. The number $a$ above the fraction bar is the numerator, and $b$ below it is the denominator. The denominator can never be 0.
• Simplifying and expanding: you can divide (or multiply) the numerator and the denominator by the same number without changing the value: $\frac{6}{8} = \frac{3}{4}$.
• Adding and subtracting fractions: first write the fractions with a common denominator, then add the numerators: $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$.
• Multiplication: numerator times numerator and denominator times denominator: $\frac{2}{3}\cdot\frac{3}{5} = \frac{6}{15} = \frac{2}{5}$.
• Division: multiply by the reciprocal (the fraction turned upside down): $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b}\cdot\frac{d}{c}$.
• Percent means hundredths: $p\% = \frac{p}{100}$. So $25\% = 0.25 = \frac{1}{4}$.
• Ratio: $1 : 2 : 3$ means 1 part, 2 parts and 3 parts, making 6 equal parts in total.
• Growth factor: an increase of $p\%$ is the same as multiplying by $1 + \frac{p}{100}$, and a decrease of $p\%$ means multiplying by $1 - \frac{p}{100}$.

How to solve the problems

1. Look for parentheses and evaluate them first. Then powers, then multiplication and division, and finally addition and subtraction.
2. Put negative numbers in parentheses when you substitute them into an expression, for example $3\cdot(-2)$.
3. Fractions: find a common denominator for addition and subtraction, multiply straight across for multiplication, and flip the second fraction for division. Simplify at the end.
4. Percent: write the percentage as a decimal or as a growth factor, and multiply. Several changes in a row give several growth factors that are multiplied together.
5. Check the answer: is the size reasonable? A discount, for example, must give a lower price.

Example

Calculate $\frac{2}{3} + \frac{1}{4}\cdot 2$.

Multiplication first: $\frac{1}{4}\cdot 2 = \frac{2}{4} = \frac{1}{2}$. The common denominator is 6:

An electric motor costs NOK 8000. The price is raised by 25%, and later it is reduced by 20%. The growth factors are $1.25$ and $0.80$:

So the price is back where it started, even though 25 and 20 are different numbers. The reason is that the 20 percent is taken of a larger amount than the 25 percent was.

Common mistakes

• Working from left to right without thinking about the order: $2 + 3\cdot 4$ is 14, not 20.
• Believing that $-3^2 = 9$. The power applies only to the number 3, so $-3^2 = -9$, while $(-3)^2 = 9$.
• Adding the denominators: $\frac{1}{2} + \frac{1}{3}$ is not $\frac{2}{5}$.
• Believing that +20% followed by −20% gives the original value. The correct result is $1.2\cdot 0.8 = 0.96$, i.e. 4% lower.
• Mixing up percent and percentage points. When an interest rate goes from 2% to 3%, it has increased by 1 percentage point, but by 50%.
• Dividing by the wrong number in a percentage change. Always divide by the old value.

Concepts in this part

Practise numbers, fractions and percent in the app →

2. Powers, roots and scientific notation

What is it about?

Powers are a shorthand for repeated multiplication: instead of writing $5\cdot5\cdot5\cdot5$ we write $5^4$. Engineers use powers all the time - in formulas for area, volume, energy and electrical power, and whenever quantities grow or shrink quickly. Roots are the opposite of powers, and powers of ten give a compact way of writing very large or very small numbers, such as the distance to a star or the diameter of an atom.

This unit builds up the rules for powers and roots from scratch, and shows how they are used to work with scientific notation and unit prefixes such as kilo, milli and micro.

Concepts and formulas

• Power: $a^n = a\cdot a\cdots a$ ($n$ factors). $a$ is called the base, $n$ the exponent.
• Power rules (same base $a\neq0$):

• $(ab)^n=a^nb^n$ and $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$.
• Special exponents: $a^0=1$ (for $a\neq0$), and $a^{-n}=\frac{1}{a^n}$ (a negative exponent flips the fraction).
• Root: $\sqrt[n]{a}$ is the number that, raised to the $n$-th power, gives $a$. A square root is written simply $\sqrt{a}$ ($n=2$). As a fractional exponent: $a^{1/n}=\sqrt[n]{a}$ and $a^{m/n}=\left(\sqrt[n]{a}\right)^m$.
• Rules for roots: $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ and $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$, but $\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$.
• Scientific notation: a number is written as $t=a\cdot10^n$, where $1\le|a|<10$ and $n$ is an integer. Large numbers give a positive $n$, numbers between 0 and 1 give a negative $n$.
• Unit prefixes are built on powers of ten: giga (G) $=10^9$, mega (M) $=10^6$, kilo (k) $=10^3$, milli (m) $=10^{-3}$, micro ($\mu$) $=10^{-6}$, nano (n) $=10^{-9}$.

How to solve the problems

1. Same base: add the exponents for multiplication, subtract for division, multiply the exponents for a power of a power.
2. Negative exponent: rewrite as a fraction with a positive exponent in the denominator.
3. Root: rewrite as a fractional exponent when that makes the calculation easier, or find the root directly if the number is a known square or cube.
4. Powers of ten: handle the numbers in front (the mantissas) separately, and add or subtract the exponents of the powers of ten separately.
5. Write the answer in scientific notation if asked to, making sure the mantissa is between 1 and 10.
6. Check the order of magnitude at the end: is the answer a reasonable size?

Example

Calculate $\dfrac{(2.0\cdot10^5)\cdot(3.0\cdot10^{-2})}{4.0\cdot10^3}$.

Handle the mantissas separately: $\dfrac{2.0\cdot3.0}{4.0}=1.5$. Handle the powers of ten separately: $\dfrac{10^5\cdot10^{-2}}{10^3}=10^{5-2-3}=10^0=1$. The answer is

Also find $\sqrt[3]{64}\cdot 4^{-1}$. The cube root of 64 is 4 (since $4^3=64$), and $4^{-1}=\frac14$, so the answer is $4\cdot\frac14=1$.

Common mistakes

• Believing $a^m\cdot a^n=a^{mn}$. The correct rule is to add the exponents: $a^{m+n}$.
• Confusing $a^{-n}$ with $-a^n$. A minus sign in the exponent means "flip the fraction", not "make the number negative": $2^{-3}=\frac18$, not $-8$.
• Believing $\sqrt{a^2}=a$ for every $a$. The correct result is $\sqrt{a^2}=|a|$, which is positive even when $a$ is negative.
• Adding inside the root sign: $\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$.
• The wrong sign on the exponent in scientific notation, for example writing $4.5\cdot10^4$ instead of $4.5\cdot10^{-4}$ for a small number.
• Forgetting that the mantissa in scientific notation must be between 1 and 10 (not, for example, $45\cdot10^{-5}$).

Concepts in this part

Practise powers, roots and scientific notation in the app →

3. Algebra and equations

What is it about?

In many engineering formulas there is an unknown quantity to find - that is when we use letters (variables) and solve an equation. Algebra is the language we use to calculate with unknown quantities instead of just numbers, and it gives us a fixed recipe that works no matter which numbers appear in the formula. Being able to work with algebraic expressions, solve equations and rearrange formulas is one of the most important skills in engineering - you will use it in almost every course.

Concepts and formulas

• Expression, term, coefficient: in the expression $3x^2-5x+2$, the terms are $3x^2$, $-5x$ and $2$, and 3 and -5 are the coefficients in front of x.
• Multiplying into a parenthesis (the distributive law): $a(b+c)=ab+ac$. A minus sign in front of a parenthesis flips the sign of every term: $-(a-b)=-a+b$.
• The special product formulas:

• Factoring: writing an expression as a product, the reverse of expanding. Look for a common factor: $6x^2-9x=3x(2x-3)$.
• Equation: a statement that two expressions are equal, true for particular values of x. Solving an equation means finding those values.
• The balance principle: you can apply the same operation (+, -, ×, ÷ by a number $\neq0$) to both sides of the equals sign without changing the solution.
• Quadratic equation $ax^2+bx+c=0$ ($a\neq0$) is solved with the quadratic formula:

The expression under the root sign, $b^2-4ac$, is called the discriminant. If it is positive, there are two solutions; if it is zero, one (double) solution; if it is negative, no real solutions.

• Two equations with two unknowns (a linear system) are solved by substitution or elimination: you eliminate one of the unknowns.

How to solve the problems

1. Simplify each side separately: expand any parentheses and combine like terms.
2. Move all the x-terms to one side and the numbers to the other (flip the sign when a term crosses the equals sign).
3. Divide by the coefficient in front of x at the end.
4. For quadratic equations: get everything into the form $ax^2+bx+c=0$ and substitute into the quadratic formula. Calculate the discriminant first.
5. To isolate a variable in a formula: treat it as the unknown x, and perform exactly the same operations.
6. Check the answer by substituting it back into the original equation.

Example

Solve the equation $3(x-2)=2x+4$.

Expand the parenthesis: $3x-6=2x+4$. Move $2x$ over: $3x-2x-6=4$, i.e. $x-6=4$. Add 6 to both sides: $x=10$.

Also solve $x^2-2x-8=0$. Here $a=1,\,b=-2,\,c=-8$. The discriminant is $b^2-4ac=(-2)^2-4\cdot1\cdot(-8)=4+32=36$. Then

which gives $x=4$ or $x=-2$.

Common mistakes

• Forgetting to multiply out every term in a parenthesis, especially with a minus sign in front: $-(x-3)=-x+3$, not $-x-3$.
• Believing $(a+b)^2=a^2+b^2$. The term $2ab$ is missing.
• Sign errors when a term is moved across the equals sign.
• Dividing only one side of the equation, or dividing by an expression that could be zero.
• Swapping $b$ and $-b$, or forgetting the $\pm$ in the quadratic formula.
• Believing quadratic equations always have two solutions - check the discriminant.

Concepts in this part

Practise algebra and equations in the app →

4. Functions and graphs

What is it about?

A function is a rule that gives exactly one output value for each input value. We write $f(x)$ for the value the function gives at $x$. Engineers use functions to model relationships between quantities - how distance depends on time, how cost depends on quantity, how voltage depends on current. The graph of a function is a picture of this relationship, and being able to read and build graphs is central to understanding what a formula actually says.

This unit covers the two most important function types in this course: linear functions (straight lines) and quadratic functions (parabolas), and how to read information out of their graphs.

Concepts and formulas

• Function $f(x)$: for every x-value (the domain) the function gives exactly one y-value ($f(x)$, the range).
• Linear function: $f(x)=ax+b$, the graph is a straight line. $a$ is the slope (the change in y per change in x), and $b$ is the y-intercept (where the line crosses the y-axis, at $x=0$).
• Slope between two points: $a=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\Delta y}{\Delta x}$.
• A positive $a$ gives a rising graph, a negative $a$ a falling graph, and $a=0$ a horizontal line. A larger $|a|$ gives a steeper line.
• A line through a point with a known slope: $y-y_1=a(x-x_1)$.
• Quadratic function: $f(x)=ax^2+bx+c$ ($a\neq0$), the graph is a parabola. $a>0$: the parabola opens upward (has a minimum); $a<0$: it opens downward (has a maximum).
• Vertex: the x-coordinate is $x_t=-\dfrac{b}{2a}$; substitute it back into the function to find the y-coordinate.
• Zeros (where the graph crosses the x-axis): solve $ax^2+bx+c=0$ with the quadratic formula (from the previous unit).
• The axis of symmetry of the parabola is the vertical line $x=x_t$ through the vertex.

How to solve the problems

1. To find the slope between two points: substitute into $a=\Delta y/\Delta x$, being careful with the order.
2. To find the function expression of a straight line: find $a$ first, then substitute one point to find $b$.
3. To read a graph: look at where it crosses the axes, whether it rises or falls, and (for parabolas) where the vertex is.
4. To find the vertex of a parabola: use $x_t=-b/2a$, then substitute it into the function for the y-value.
5. To find the zeros: set $f(x)=0$ and solve the quadratic equation.
6. Check the answer: substitute the x-value back into the function expression and see that you get the right y-value.

Example

A straight line passes through the points $(1,3)$ and $(4,12)$. Find the function expression.

Slope: $a=\dfrac{12-3}{4-1}=\dfrac{9}{3}=3$. Substitute the point $(1,3)$: $3=3\cdot1+b \Rightarrow b=0$. The function is $f(x)=3x$.

Find the vertex of $f(x)=-2x^2+8x+3$. Here $a=-2,b=8$, so $x_t=-\dfrac{8}{2\cdot(-2)}=2$. Then $f(2)=-2\cdot4+16+3=11$. The vertex is $(2,11)$, and since $a<0$ this is a maximum.

Common mistakes

• Swapping $\Delta y$ and $\Delta x$ in the slope formula (remember: change in y over change in x).
• Believing that $b$ in $f(x)=ax+b$ is the slope - it is $a$ that is the slope.
• Sign errors in $x_t=-b/2a$, especially when $b$ or $a$ is negative.
• Believing every quadratic function has zeros - if the vertex lies above the x-axis (parabola opening downward) or below the x-axis (parabola opening upward), there may be no zeros. This connects to the sign of the discriminant.
• Reading the graph the wrong way round (mixing up the x- and y-axis) or misreading the scale.

Concepts in this part

Practise functions and graphs in the app →

5. Exponential and logarithmic functions

What is it about?

Many quantities in engineering and nature change by the same factor in each time step, not by the same amount. A bank deposit at 4% interest is multiplied by 1.04 every year. A radioactive source loses the same fraction every year. A discharging capacitor, a cooling cup of coffee and a growing bacterial culture all follow the same pattern. This is called exponential growth or decay.

Compare this with linear growth, where we add the same amount each time ($+3$, $+3$, $+3$, ...). In exponential growth we multiply by the same number ($\cdot 2$, $\cdot 2$, $\cdot 2$, ...). Exponential growth looks slow at first, but eventually becomes enormously fast.

The logarithm is the reverse of a power. It answers the question "which exponent do I need?". You need it whenever the unknown sits in the exponent, for example when you want to find how long something takes.

Concepts and formulas

• Exponential function: $f(x) = a\cdot b^x$, where $a = f(0)$ is the starting value and $b > 0$ is the growth factor.
• $b > 1$ gives growth, and $0 < b < 1$ gives decay.
• Percentage change: an increase of $p$% gives $b = 1 + \frac{p}{100}$, and a decrease of $p$% gives $b = 1 - \frac{p}{100}$. Example: $+5$% gives $1.05$, and $-20$% gives $0.80$.
• The number $e \approx 2.718$ is the "natural" base. In physics and calculus we almost always write $N(t) = N_0e^{kt}$, where $k > 0$ means growth and $k < 0$ means decay. Any base can be rewritten: $b^t = e^{t\ln b}$.
• Logarithms: $\lg x = y$ means $10^y = x$, and $\ln x = y$ means $e^y = x$. In general, $\log_b x = y$ means $b^y = x$, and $\log_b x = \frac{\ln x}{\ln b}$. (Many English texts write $\log$ or $\log_{10}$ for $\lg$.)
• The logarithm is only defined for $x > 0$. Also, $\ln 1 = 0$ and $\ln e = 1$.
• $e^{\ln x} = x$ and $\ln(e^x) = x$: the two functions undo each other.
• Rules (they also hold for $\lg$):

• Half-life and doubling time: $T = \frac{\ln 2}{|k|}$. After a time $t$, the fraction $\left(\tfrac12\right)^{t/T}$ is left.

How to solve the problems

1. Find the starting value ($a$ or $N_0$) and the growth factor ($b$) or the rate ($k$).
2. Write the model, for example $N(t) = N_0\cdot b^t$.
3. To find a value: insert $t$ and calculate.
4. To find a time (the unknown is in the exponent): isolate the power on one side, take $\ln$ of both sides, use $\ln(b^t) = t\ln b$ and divide.
5. Check that the answer is reasonable: decay should give smaller values, and a time should be positive.

Example

The activity of a radioactive source drops by 12% per year. How long does it take for the activity to be halved?

1. The growth factor is $b = 1 - 0.12 = 0.88$, so $A(t) = A_0\cdot 0.88^t$.
2. Halved means $A(t) = 0.5A_0$. Divide by $A_0$: $0.88^t = 0.5$.
3. Take $\ln$ of both sides: $t\ln 0.88 = \ln 0.5$.
4. $t = \dfrac{\ln 0.5}{\ln 0.88} = \dfrac{-0.6931}{-0.1278} \approx 5.42$ years.

The same model with $e$: $k = \ln 0.88 \approx -0.128$, so $A(t) = A_0e^{-0.128t}$ and $T = \ln 2/0.128 \approx 5.42$ years. Both ways of writing the model give the same answer.

Common mistakes

• Treating percentages linearly: 10% growth two years in a row gives $1.1^2 = 1.21$, which is 21%, not 20%.
• The wrong factor for a decrease: a 20% decrease gives $0.8$, not $0.2$ or $-0.2$.
• $\ln(a + b)$ is not $\ln a + \ln b$. The rules apply to products, quotients and powers.
• $(e^x)^2 = e^{2x}$, not $e^{x^2}$.
• Mixing up $\lg$ (base 10) and $\ln$ (base $e$) on the calculator.
• Taking the logarithm of zero or of a negative number. It is not defined.

Concepts in this part

Practise exponential and logarithmic functions in the app →

6. Trigonometry and geometry

What is it about?

Trigonometry links angles and lengths in triangles. Engineers use it all the time: to split a force into a horizontal and a vertical part, to find the slope of a ramp or a roof, to calculate the length of a diagonal brace, or to describe something that rotates or oscillates. In calculus and physics, $\sin$ and $\cos$ appear as functions, and then angles are measured in radians.

Concepts and formulas

• The angles of a triangle add up to $180^\circ$. A right angle is $90^\circ$.
• Radians: one full turn is $360^\circ = 2\pi$ rad. Conversion: $\text{rad} = \text{degrees}\cdot\frac{\pi}{180}$. So $180^\circ = \pi$ and $90^\circ = \pi/2$.
• Pythagoras (right triangle, $c$ is the hypotenuse, the longest side): $a^2 + b^2 = c^2$. The two shorter sides are called legs.
• For an acute angle $\theta$ in a right triangle:

• If you know the sides and need the angle, use the inverse functions $\sin^{-1}$, $\cos^{-1}$ and $\tan^{-1}$ (often "shift sin" on the calculator).
• Exact values: $\sin 30^\circ = \tfrac12$, $\cos 60^\circ = \tfrac12$, $\sin 45^\circ = \cos 45^\circ = \tfrac{\sqrt2}{2}$, $\sin 60^\circ = \cos 30^\circ = \tfrac{\sqrt3}{2}$.
• The unit circle: the point at angle $\theta$ (measured counterclockwise from the positive $x$-axis) has coordinates $(\cos\theta, \sin\theta)$. This is how sin and cos are defined for all angles. It gives $\sin^2\theta + \cos^2\theta = 1$ and shows the signs: for example, $\cos\theta < 0$ when $90^\circ < \theta < 270^\circ$.
• Any triangle (sides $a, b, c$ opposite the angles $A, B, C$):

• Circle with radius $r$: circumference $2\pi r$, area $\pi r^2$, arc length $s = r\theta$ and sector area $A = \tfrac12 r^2\theta$ (with $\theta$ in radians).

How to solve the problems

1. Draw a sketch, and mark what you know and what you need to find.
2. Is there a right angle? Use Pythagoras (sides only) or $\sin$, $\cos$, $\tan$ (sides and an angle).
3. No right angle? Use the law of cosines when you know two sides and the angle between them (or all three sides), and the law of sines when you know a side and the angle opposite it.
4. Check that the calculator is in the right mode: DEG for degrees, RAD for radians.
5. Check the answer: the hypotenuse is the longest side, and the largest angle is opposite the longest side.

Example

A ramp is 4.0 m long (along the slope) and rises 0.80 m. What is the angle of the slope, and how long is the ramp measured along the ground?

1. The rise is the opposite side and the ramp is the hypotenuse: $\sin\theta = \dfrac{0.80}{4.0} = 0.20$.
2. $\theta = \sin^{-1}(0.20) \approx 11.5^\circ$.
3. The ground length is the adjacent side: $b = \sqrt{4.0^2 - 0.80^2} = \sqrt{15.36} \approx 3.92$ m.

Without a right angle: two bars of 3.0 m and 5.0 m are joined at one point with $60^\circ$ between them. The distance between their free ends is $c = \sqrt{3^2 + 5^2 - 2\cdot 3\cdot 5\cos 60^\circ} = \sqrt{19} \approx 4.36$ m.

Common mistakes

• The calculator is in RAD while the angle is in degrees (or the other way round). $\sin 30$ in radians gives $-0.988$, not $0.5$.
• Swapping the opposite and the adjacent side. Both are defined relative to the angle you are looking at.
• Using Pythagoras or SOH CAH TOA in a triangle without a right angle.
• Thinking that $\sin^{-1} x$ means $1/\sin x$. It is the inverse function, which returns an angle.
• Using $s = r\theta$ with $\theta$ in degrees.
• The law of sines can give two possible angles ($\theta$ and $180^\circ - \theta$), because they have the same sine. Check which one fits the figure.

Concepts in this part

Practise trigonometry and geometry in the app →

7. Vectors

What is it about?

Some quantities are completely described by a single number: mass (5 kg), temperature (20 °C) and time (3 s). They are called scalars. Other quantities also have a direction: a force of 200 N pulling upward is something quite different from 200 N pulling sideways. Such quantities are called vectors. Forces, velocity, acceleration, displacement and moment are vectors.

In statics and physics you almost always need to add several forces or velocities. You do this by splitting each vector into components along the $x$- and $y$-axes, adding each direction separately and putting the result back together.

Concepts and formulas

• A vector is drawn as an arrow and written $\vec a$. Its length (magnitude) is written $|\vec a|$.
• Component form: $\vec a = (a_x, a_y)$ or $\vec a = a_x\vec i + a_y\vec j$, where $\vec i$ and $\vec j$ are unit vectors (length 1) along the axes. In three dimensions, $\vec a = (a_x, a_y, a_z)$. Some books write $\langle a_x, a_y\rangle$ or $[a_x, a_y]$.
• Magnitude: $|\vec a| = \sqrt{a_x^2 + a_y^2}$, and in 3D $|\vec a| = \sqrt{a_x^2 + a_y^2 + a_z^2}$.
• From magnitude and angle $\theta$ (measured counterclockwise from the positive $x$-axis): $a_x = |\vec a|\cos\theta$ and $a_y = |\vec a|\sin\theta$.
• Direction from components: $\tan\theta = a_y/a_x$. Check the quadrant from the signs!
• Sum: $\vec a + \vec b = (a_x + b_x, a_y + b_y)$. Number times vector: $k\vec a = (ka_x, ka_y)$.
• Vector between two points: $\overrightarrow{AB} = (x_B - x_A, y_B - y_A)$, that is, "end minus start".
• The scalar product (dot product) gives a number:

• $\vec a\cdot\vec b = 0$ means that the vectors are perpendicular. Work is a dot product: $W = \vec F\cdot\vec s$.
• Moment of a force $\vec F$ about a point, where $\vec r = (r_x, r_y)$ is the vector from the point to where the force acts: $M = r_xF_y - r_yF_x$ (positive counterclockwise). This is the $z$-component of the cross product $\vec r\times\vec F$.

How to solve the problems

To find the sum (resultant) of several forces:

1. Draw the forces and choose axes. Measure all angles counterclockwise from the positive $x$-axis.
2. Split each force into components: $F_x = F\cos\theta$ and $F_y = F\sin\theta$. Watch the signs.
3. Add each direction separately: $R_x = \sum F_x$ and $R_y = \sum F_y$.
4. Magnitude: $R = \sqrt{R_x^2 + R_y^2}$. Direction: $\theta = \tan^{-1}(R_y/R_x)$, adjusted to the correct quadrant.
5. Equilibrium means that the resultant is zero: $\sum F_x = 0$ and $\sum F_y = 0$.

Example

Two ropes pull on a hook: $F_1 = 400$ N at an angle of $30^\circ$ and $F_2 = 250$ N at $120^\circ$ (both measured from the positive $x$-axis). Find the resultant.

1. $F_{1x} = 400\cos 30^\circ \approx 346.4$ N and $F_{1y} = 400\sin 30^\circ = 200.0$ N.
2. $F_{2x} = 250\cos 120^\circ = -125.0$ N and $F_{2y} = 250\sin 120^\circ \approx 216.5$ N.
3. $R_x = 346.4 - 125.0 = 221.4$ N and $R_y = 200.0 + 216.5 = 416.5$ N.
4. $R = \sqrt{221.4^2 + 416.5^2} \approx 471.7$ N and $\theta = \tan^{-1}(416.5/221.4) \approx 62.0^\circ$.

Answering $400 + 250 = 650$ N would be wrong: forces add as vectors, not as plain numbers.

Common mistakes

• Adding the magnitudes instead of the components. Usually $|\vec a + \vec b|$ is smaller than $|\vec a| + |\vec b|$.
• Wrong signs on the components in the second, third and fourth quadrants.
• Trusting $\tan^{-1}$ blindly: the calculator only returns angles between $-90^\circ$ and $90^\circ$. For $\vec a = (-3, -3)$ the direction is $225^\circ$, not $45^\circ$.
• Using $\sin$ for the $x$-component when the angle is measured from the $x$-axis.
• $\overrightarrow{AB}$ is $B$ minus $A$, not $A$ minus $B$.
• Thinking that the dot product gives a vector. It gives a number.

Concepts in this part

Practise vectors in the app →

8. Introduction to derivatives

What is it about?

The derivative measures how steep a graph is at a point, that is, how fast something changes right there. The speed of a car is the derivative of its position. Power is the derivative of energy. This is the foundation of calculus, and you only need a few rules to get started.

Concepts and formulas

• The average rate of change from $x = a$ to $x = b$ is the slope of the line between the two points:

• The instantaneous rate of change $f'(a)$ is the slope of the tangent at the point. You get it by letting $b$ approach $a$.
• Three rules cover all polynomials:

• The derivative of a constant is zero: a horizontal line has slope 0.
• The sign tells you how the graph behaves: $f'(x) > 0$ means $f$ increases, $f'(x) < 0$ means $f$ decreases, and $f'(x) = 0$ means a horizontal tangent (possibly a maximum or minimum).
• The tangent at $x = a$: $y = f(a) + f'(a)\,(x - a)$.

How to solve the problems

1. Differentiate term by term: bring the exponent down in front and subtract 1 from the exponent.
2. Insert the $x$-value in the derivative, not in the original function.
3. Think about what the answer means: speed, slope, growth per year, etc.

Example

The position of a cart is $s(t) = 4t^2$ meters after $t$ seconds. What is its speed after 3 s?

1. The speed is the derivative of the position: $v(t) = s'(t) = 4\cdot 2t = 8t$.
2. $v(3) = 8\cdot 3 = 24$ m/s.
3. For comparison, the average speed during the first three seconds is $s(3)/3 = 36/3 = 12$ m/s.

Common mistakes

• Forgetting to subtract 1 from the exponent: $(x^3)' = 3x^2$, not $3x^3$.
• Inserting into $f$ instead of $f'$.
• Thinking that the derivative of a constant term is the constant itself. It is 0.

Concepts in this part

Practise introduction to derivatives in the app →

9. Statistics and probability

What is it about?

Engineers work with measurements that vary and with risk: how often does a component fail, what is a typical value, how much do the measurements spread? Here you learn the simplest tools for describing data and calculating with probability.

Concepts and formulas

• Mean: the sum divided by the count, $\bar x = \dfrac{\sum x_i}{n}$.
• Median: the middle value when the numbers are sorted. With an even count, take the average of the two middle ones.
• Mode: the most frequent value. Range: largest minus smallest value.
• Probability when all outcomes are equally likely:

• Complement: $P(\text{not } A) = 1 - P(A)$.
• Independent events: $P(A \text{ and } B) = P(A)\cdot P(B)$.
• At least once is easiest via the complement: $P(\text{at least one}) = 1 - P(\text{none})$.

How to solve the problems

1. Describing data: sort the numbers first, then the median and range are easy to find.
2. Probability: count possible and favourable outcomes, or multiply probabilities of independent events.
3. "At least one": compute the probability of "none" and subtract from 1.

Example

What is the probability of at least one six in three dice throws?

1. The probability of not getting a six in one throw is $5/6$.
2. No sixes in three throws: $(5/6)^3 \approx 0.579$.
3. At least one six: $1 - 0.579 = 0.421$, about 42 %.

Common mistakes

• Adding probabilities that should be multiplied: $3\cdot\tfrac16 = 0.5$ is wrong for "at least one six in three throws".
• Finding the median without sorting the numbers first.
• Letting one extreme value dominate the mean. Then the median describes the typical value better.

Concepts in this part

Practise statistics and probability in the app →