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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).

Chapter 12 · pp. 507–588

Chapter 8: Further Applications of Trigonometry

This chapter extends trigonometry beyond right triangles. It starts with the Law of Sines and the Law of Cosines for any triangle. It then moves to polar coordinates, complex numbers in polar form with DeMoivre's Theorem, vectors and the dot product, and parametric equations. This first part covers the two laws for non-right triangles.

The Law of Sines

Drop an altitude $h$ in any triangle to split it into two right triangles. One gives $h=b\sin(\alpha)$ and the other gives $h=a\sin(\beta)$ . Since both equal $h$ , we get $b\sin(\alpha)=a\sin(\beta)$ . Dividing both sides by $ab$ gives $\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}$ . Dropping the altitude from another vertex gives the third ratio. Together:

$\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$

Each side is opposite the angle with the matching letter (p. 508).

The side opposite an angle is its corresponding side, and the angle opposite a side is its corresponding angle. So $a$ pairs with $\alpha$ , $b$ with $\beta$ , and $c$ with $\gamma$ . You can set any two of the three ratios equal to make an equation. The law is most useful when you know one complete side/angle pair plus one more piece of information. A 50° angle opposite a side of length 10 is a usable pair (p. 508).

Solving for Sides

Suppose two angles and one side are known. First use the 180° angle sum to find the third angle. Then apply the Law of Sines to each unknown side.

Example: the side 10 is opposite 50°, and another angle is 30°. The third angle is $180°-50°-30°=100°$ . Then

$b=\frac{10\sin(30°)}{\sin(50°)}\approx 6.527,\qquad c=\frac{10\sin(100°)}{\sin(50°)}\approx 12.856$

(p. 509)

The same idea handles height problems in non-right triangles. Two stations 20 miles apart see an object at elevation angles 35° and 15°. The third angle is $180°-35°-15°=130°$ , and it is opposite the 20-mile side. The Law of Sines gives the slanted distance $a=\frac{20\sin(35°)}{\sin(130°)}\approx14.975$ miles. Then a right triangle gives the height: $h=a\sin(15°)\approx3.876$ miles (pp. 509–510).

Solving for Angles and the Ambiguous Case

To find an unknown angle, choose a ratio in which three of the four values are known. Isolate the sine, then use the inverse sine. The inverse sine gives an acute angle, but its supplement (180° minus that angle) has the same sine. So there are two candidates to check (pp. 510–511).

Example: 85° is opposite a side of 12, and a side of 9 is opposite $\beta$ . Then $\sin(\beta)=\frac{9\sin(85°)}{12}$ , so $\beta\approx48.34°$ or $\beta\approx131.66°$ . The second option fails because $85°+131.66°>180°$ . So $\beta\approx48.34°$ , the third angle is about $46.66°$ , and the remaining side is about $8.76$ (pp. 510–511).

This is the ambiguous case. It shows up when you know two sides and an angle that is not between them. It can give 0, 1, or 2 triangles. A careful picture or the context of the problem can also decide which angle is intended (p. 511).

No triangle. The sine can never be greater than 1 or less than −1. Take a side of 4 opposite 50° and another side of 10. Then $\sin(\theta)=\frac{10\sin(50°)}{4}\approx1.915$ , which is impossible. No triangle fits these measurements (p. 511).

Choosing between two angles. Add the obtuse candidate to the known angle. If the sum reaches 180° or more, reject it. Take a side of 6 opposite 50° and another side of 4. Then $\sin(\theta)\approx0.511$ , which gives $30.71°$ or $149.29°$ . The obtuse choice would leave $180°-50°-149.29°=-19.29°$ for the third angle, which is impossible. The only triangle has angles 50°, 30.71°, and 99.29°, and its third side is $c\approx7.730$ (p. 512).

The Law of Cosines

The Law of Sines needs a known side/angle pair. It fails when you know two sides and the angle between them, because that angle is not opposite either known side. For example, a boat travels 10 miles, turns 20°, and travels 8 miles more. For this case we need a new tool (p. 512).

Drop an altitude to split side $b$ into pieces $x$ and $b-x$ , where $x=c\cos(\alpha)$ . Write the Pythagorean Theorem for each right triangle and set the two expressions for $h^2$ equal. Simplifying gives $a^2=b^2+c^2-2bc\cos(\alpha)$ . By symmetry:

$a^2=b^2+c^2-2bc\cos(\alpha)$ $b^2=a^2+c^2-2ac\cos(\beta)$ $c^2=a^2+b^2-2ab\cos(\gamma)$

The side on the left is opposite the angle in the cosine. The right side uses the two sides next to that angle (pp. 512–513).

If the angle is 90°, then $\cos(90°)=0$ . The formula becomes $c^2=a^2+b^2$ , the Pythagorean Theorem. That is why the Law of Cosines is called the Generalized Pythagorean Theorem. The extra term $-2ab\cos(\gamma)$ corrects for an angle that is not right. For an obtuse angle the cosine is negative, so $c^2$ is larger than $a^2+b^2$ (p. 513).

Using the Law of Cosines

Finding a side. When a path turns by some angle, the interior angle of the triangle is 180° minus the turn. For the boat, the interior angle is $180°-20°=160°$ :

$x^2=8^2+10^2-2(8)(10)\cos(160°)\approx314.35$

So $x\approx17.73$ miles from port (p. 514).

Finding the other angles. After you find the third side, you have a side/angle pair. Use the Law of Sines for the rest. Take sides 10 and 12 around a 30° angle. Then $x^2=244-120\sqrt3$ , so $x\approx6.013$ . Next, $\sin(\theta)=\frac{10\sin(30°)}{6.013}$ gives $\theta\approx56.256°$ . The last angle is $180°-30°-56.256°\approx93.744°$ . The other candidate, 123.744°, is rejected because the figure shows an acute angle (pp. 514–515).

Three sides known. The Law of Cosines gives an angle directly. For sides 18, 25, and 20, take the angle $\alpha$ opposite 20. Then $20^2=18^2+25^2-2(18)(25)\cos(\alpha)$ , so $\cos(\alpha)=\frac{549}{900}$ and $\alpha\approx52.41°$ . The inverse cosine returns one angle between 0° and 180°, which covers every angle a triangle can have. So there is never an ambiguous case here (p. 515).

Applications: Triangulation and Heights

Triangulation. Two towers stand 6000 ft apart on an east-west highway. You are north of the highway, 5050 ft from the first tower and 2420 ft from the second. The Law of Cosines gives the angle at the first tower, $\theta\approx23.3°$ . Right-triangle trigonometry then gives your position: $x=5050\cos(\theta)\approx4637$ ft east and $y=5050\sin(\theta)\approx2000$ ft north. If you did not know which side of the highway you were on, there would be two possible locations. A third tower removes that ambiguity (p. 516).

Height from two elevation angles. A hill is seen at 24°, and then at 22° after stepping back 200 ft. The triangle with the 200 ft side is not a right triangle, so start there. Its obtuse angle is $180°-24°=156°$ , which makes the remaining angle $180°-156°-22°=2°$ . The Law of Sines gives the long side $x=\frac{200\sin(22°)}{\sin(2°)}\approx2146.77$ ft. Then use the right triangle with the 24° angle: $h=x\sin(24°)\approx873$ ft (p. 517).

Locating points with polar coordinates

A Cartesian point is described by how far it is across and up. A polar point is described by how far it is from the origin and which way you turn to face it. A polar pair $(r,\theta)$ has two parts. The distance $r$ is measured from the origin. The angle $\theta$ is measured in standard position, starting at the positive horizontal axis. A polar grid is made of circles at equal radii and rays at equal angles (p. 524).

Example. To plot $\left(3,\frac{5\pi}{6}\right)$ , turn to the ray at $\frac{5\pi}{6}$ , which is in quadrant II. Then go 3 units out along it (p. 524).

Negative $r$ . A negative $r$ works like a negative number on a number line: you go the opposite way. First rotate to $\theta$ , then move $|r|$ units backward through the origin. For $\left(-2,\frac{\pi}{4}\right)$ , the ray at $\frac{\pi}{4}$ points into quadrant I, so the point lands in quadrant III, 2 units from the origin (pp. 524–525).

Many names for one point. The point above is also $\left(2,\frac{5\pi}{4}\right)$ . Adding or subtracting any number of full turns ( $2\pi$ ) to $\theta$ gives the same location, so $\left(2,\frac{13\pi}{4}\right)$ names it too. Every point therefore has infinitely many polar representations. Cartesian coordinates do not have this feature, because each point has just one $(x,y)$ pair (p. 525).

Converting points between systems

Drop a perpendicular from the point to the horizontal axis. This makes a right triangle with hypotenuse $r$ , so the relationships from earlier trigonometry apply:

$x=r\cos(\theta),\quad y=r\sin(\theta),\quad \tan(\theta)=\frac{y}{x},\quad x^2+y^2=r^2$

Knowing the unit circle values makes these conversions quick (p. 525).

Polar to Cartesian. Just substitute. For $(r,\theta)=\left(5,\frac{2\pi}{3}\right)$ :

$x=5\cos\left(\tfrac{2\pi}{3}\right)=-\tfrac52,\qquad y=5\sin\left(\tfrac{2\pi}{3}\right)=\tfrac{5\sqrt3}{2}$

The Cartesian point is $\left(-\frac52,\frac{5\sqrt3}{2}\right)$ (p. 526).

Cartesian to polar. This takes more care, because an inverse trig function gives only one of two possible angles.

Find $r=\sqrt{x^2+y^2}$ .
Use an inverse trig function to get an angle.
Check which quadrant the point is in, and use the symmetric angle if needed.

For $(-3,-4)$ , $r=\sqrt{9+16}=5$ . Then $\cos^{-1}\left(-\frac35\right)\approx 2.214$ , which is in quadrant II. The point is in quadrant III, so use $2\pi-2.214\approx 4.069$ . The polar point is $(5,\,4.069)$ (p. 526).

Graphing polar equations

A polar equation links $r$ and $\theta$ , just as a Cartesian equation links $x$ and $y$ . To sketch one, make a table of $\theta$ values and matching $r$ values, plot the points, and connect them. Most polar graphs are drawn on a Cartesian grid. A graphing calculator must be in polar mode, where you enter $r$ as a function of $\theta$ (pp. 527–528).

Spiral: $r=\theta$ . Here $r$ grows as $\theta$ grows, so the point turns around the origin while moving steadily outward. The result is a spiral. Even though $r$ is a function of $\theta$ , the curve fails the vertical line test. It cannot be written as $y=f(x)$ (p. 527).

Circle: $r=3$ . The angle does not appear, so any angle is allowed. Every point is exactly 3 units from the origin. This is like $y=3$ in Cartesian coordinates, where $x$ is unrestricted and the graph is a horizontal line. Here the graph is a circle centered at the origin (p. 527).

One cycle of $r=4\cos(\theta)$ . This graph is a circle through the origin. It starts at $(4,0)$ when $\theta=0$ . To find when it first returns there, set the $x$ -value $r\cos(\theta)$ equal to 4. This gives $4\cos^2(\theta)=4$ , so $\cos^2(\theta)=1$ and $\cos(\theta)=\pm1$ . The solutions are $\theta=0$ and $\theta=\pi$ . So $0\le\theta\le\pi$ traces the whole circle once (p. 528).

Limaçon: $r=4\sin(\theta)+2$ . The inner loop forms where $r$ is negative. It starts and ends at the origin, where $r=0$ . Solve $4\sin(\theta)+2=0$ to get $\sin(\theta)=-\frac12$ . On $[0,2\pi)$ this gives $\theta=\frac{7\pi}{6}$ and $\theta=\frac{11\pi}{6}$ . The inner loop is traced for $\frac{7\pi}{6}\le\theta\le\frac{11\pi}{6}$ (p. 529).

Rose: $r=\cos(3\theta)$ . This graph is a 3-leaf rose, and one full cycle takes $[0,\pi)$ . To find one loop, solve $\cos(3\theta)=0$ . Let $u=3\theta$ , so $u=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}$ . Dividing by 3 gives $\theta=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$ . The first two give one loop: $\frac{\pi}{6}\le\theta\le\frac{\pi}{2}$ .

On that interval $r$ is negative. At $\theta=\frac{\pi}{3}$ , $r=\cos(\pi)=-1$ . A first-quadrant angle with negative $r$ plots in quadrant III, so this loop sits in the third quadrant. The graph starts at $(1,0)$ when $\theta=0$ (pp. 529–530).

Converting equations between systems

The same four relationships convert whole equations. Not every equation converts into a neat form, but sometimes the other form is more useful (pp. 530–531).

Cartesian to polar. Replace $x$ with $r\cos(\theta)$ , $y$ with $r\sin(\theta)$ , and $x^2+y^2$ with $r^2$ . Then isolate $r$ if you can.

For $x^2+y^2=6y$ : write $r^2=6r\sin(\theta)$ . Move terms and factor: $r\big(r-6\sin(\theta)\big)=0$ . The solution $r=0$ is only the origin, which the other factor already covers. So $r=6\sin(\theta)$ , a circle that sits on the origin and reaches up to $(0,6)$ (p. 531).
For $y=3x+2$ : write $r\sin(\theta)=3r\cos(\theta)+2$ . Collect the $r$ terms, factor, and divide to get $r=\dfrac{2}{\sin(\theta)-3\cos(\theta)}$ . This is clumsier than the original, but still valid (p. 531).

Polar to Cartesian. Clear any fraction first. Then use $\cos(\theta)=\frac{x}{r}$ , $\sin(\theta)=\frac{y}{r}$ , and $r^2=x^2+y^2$ until only $x$ and $y$ are left.

For $r=\dfrac{3}{1-2\cos(\theta)}$ : clear the fraction to get $r-2r\cos(\theta)=3$ . Substitute $r\cos(\theta)=x$ to get $r-2x=3$ , so $r=3+2x$ . Square both sides and replace $r^2$ : $x^2+y^2=(3+2x)^2$ . This cannot be solved into a function $y=f(x)$ (pp. 531–532).
For $r=\sin(2\theta)$ : use the double angle identity $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ . Then $r=2\cdot\frac{y}{r}\cdot\frac{x}{r}$ , so $r^3=2xy$ . Replacing $r$ gives $(x^2+y^2)^{3/2}=2xy$ (p. 532).

The imaginary unit and the complex plane

The imaginary unit is $i=\sqrt{-1}$ . Any real multiple of $i$ is also called imaginary. To simplify the root of a negative number, split off the $-1$ . For example, $\sqrt{-9}=\sqrt{9}\cdot\sqrt{-1}=3i$ (p. 537).

A complex number has the form $z=a+bi$ . Here $a$ is the real part and $b$ is the imaginary part, and both are real numbers (p. 537).

A single number line cannot show two parts, so we cross two lines to make the complex plane. The horizontal axis is the real axis and the vertical axis is the imaginary axis. We plot $a+bi$ like the Cartesian point $(a,b)$ . So $3-4i$ is 3 units right and 4 units down (pp. 537–538). Writing $z=x+yi$ makes this link to $(x,y)$ easy to see (p. 538).

Adding, multiplying and dividing complex numbers

Addition and subtraction. Combine the real parts, then combine the imaginary parts:

$(3-4i)+(2+5i)=(3+2)+(-4+5)i=5+i$

(p. 538)

Multiplication. Expand as you would with polynomials (FOIL), then replace $i^2$ with $-1$ :

$(2-3i)(1+4i)=2+8i-3i-12i^2=2+5i+12=14+5i$

(p. 529 method, shown on p. 539)

Division. We cannot leave $i$ in a denominator. Multiply the top and bottom by the complex conjugate of the denominator, which is the same number with the sign of the imaginary part flipped. The denominator then becomes a real number. For $\frac{2+5i}{4-i}$ , use $4+i$ :

Numerator: $(2+5i)(4+i)=8+2i+20i+5i^2=3+22i$
Denominator: $(4-i)(4+i)=16-i^2=17$

So the quotient is $\frac{3}{17}+\frac{22}{17}i$ (pp. 539–540).

Polar form and Euler's Formula

Because $x+yi$ matches the point $(x,y)$ , the polar-coordinate rules apply. These are $x=r\cos\theta$ , $y=r\sin\theta$ , $\tan\theta=\frac{y}{x}$ and $x^2+y^2=r^2$ (p. 540). Substituting gives

$z=x+yi=r\cos\theta+ir\sin\theta.$

Euler's Formula says $re^{i\theta}=r\cos\theta+ir\sin\theta$ . So the compact polar form is $z=re^{i\theta}$ (p. 541). In this form, $r$ is the distance from the origin and $\theta$ is the angle from the positive real axis.

Two quick examples:

$4i$ sits 4 units up the imaginary axis, so $r=4$ and $\theta=\frac{\pi}{2}$ . Thus $4i=4e^{i\pi/2}$ .
$-8$ is written as $-8+0i$ . It lies 8 units along the negative real axis, so $r=8$ , $\theta=\pi$ and $-8=8e^{i\pi}$ . Check: $8\cos\pi+8i\sin\pi=-8$ (p. 541).

To go back to $a+bi$ form, evaluate the cosine and sine. For example, $3e^{i\pi/6}=3\cdot\frac{\sqrt3}{2}+3\cdot\frac12 i=\frac{3\sqrt3}{2}+\frac32 i$ (p. 543).

Converting $x+yi$ to polar form

Compute $r=\sqrt{x^2+y^2}$ .
Find $\theta$ from $\cos\theta=\frac{x}{r}$ or $\tan\theta=\frac{y}{x}$ .
Check the quadrant of the point and pick the angle that matches it.

Example: for $-4+4i$ , $r=\sqrt{32}=4\sqrt2$ and $\cos\theta=-\frac{\sqrt2}{2}$ . The point is in quadrant II, so $\theta=\frac{3\pi}{4}$ . Then $-4+4i=4\sqrt2\,e^{i3\pi/4}$ (p. 542).

Step 3 matters because inverse tangent only returns angles in quadrants I and IV. For $-3-5i$ , $r=\sqrt{34}$ and $\tan^{-1}\!\left(\frac53\right)\approx1.0304$ . But the point is in quadrant III, so add $\pi$ to get $\theta\approx4.172$ (p. 542).

Powers with DeMoivre's Theorem

Polar form makes powers easy because it lets us use exponent rules. To find a power:

Convert to polar form.
Raise $r$ to the power and multiply the angle by the exponent.
Convert back to $a+bi$ if needed.

DeMoivre's Theorem states this shortcut. If $z=r(\cos\theta+i\sin\theta)$ , then for any integer $n$ ,

$z^n=r^n\big(\cos(n\theta)+i\sin(n\theta)\big).$

Example: find $(-4+4i)^6$ . Since $-4+4i=4\sqrt2\,e^{i3\pi/4}$ :

$(4\sqrt2)^6=32768,\qquad 6\cdot\frac{3\pi}{4}=\frac{9\pi}{2}.$

So the result is $32768e^{i9\pi/2}$ . The angle $\frac{9\pi}{2}$ is coterminal with $\frac{\pi}{2}$ , where cosine is 0 and sine is 1. Therefore $(-4+4i)^6=32768i$ (pp. 543–544). Multiplying out six times by hand would take far longer.

Roots of complex numbers

A square root is a power of $\frac12$ , so we halve the angle and take the square root of $r$ . For $9i=9e^{i\pi/2}$ :

$(9i)^{1/2}=3e^{i\pi/4}=\frac{3\sqrt2}{2}+\frac{3\sqrt2}{2}i.$

(pp. 544–545)

This is only one root. Just as $x^2=4$ has the solutions $2$ and $-2$ , the equation $z^2=9i$ has a second solution. Likewise $z^3=8$ has three solutions, and only $z=2$ is real (p. 545).

The other roots come from the fact that an angle has many coterminal versions. Since 8 lies on the positive real axis, we can write

$8=8e^{0i}=8e^{2\pi i}=8e^{4\pi i}.$

Take the $\frac13$ power of each. The modulus becomes $8^{1/3}=2$ and the angle is divided by 3:

Angle $0$ : $2e^{0i}=2$
Angle $2\pi$ : $2e^{2\pi i/3}=2\cos\frac{2\pi}{3}+2i\sin\frac{2\pi}{3}=-1+\sqrt3\,i$
Angle $4\pi$ : $2e^{4\pi i/3}=-1-\sqrt3\,i$

These are all the solutions of $z^3=8$ (p. 547). Using a fourth angle such as $6\pi$ would only repeat the first root, so three roots are enough.

Where the roots sit

All three cube roots of 8 have modulus 2, so they lie on a circle of radius 2 centered at the origin of the complex plane. Their angles are $0$ , $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ , which are $\frac{2\pi}{3}$ apart. So the roots are equally spaced around the circle (p. 547). The same pattern holds for any $n$ roots of a number: they are evenly spread around a circle (p. 547).

Vectors and Their Geometric Arithmetic

A vector is a quantity with both a size (magnitude) and a direction. We draw it as an arrow. A vector from point $A$ to point $B$ is written $\overrightarrow{AB}$ , or with a bold letter or a letter with an arrow, such as $\vec u$ . A walk of 3 miles north is a vector, because it has a length (3 miles) and a direction (north). (p. 551)

Adding vectors. Adding vectors means doing one movement and then the next. To find $\vec u+\vec v$ , draw $\vec u$ , then draw $\vec v$ starting at the end of $\vec u$ . The sum is the arrow that goes straight from the start of $\vec u$ to the end of $\vec v$ . For example, if a woman walks 3 miles north and then 2 miles southeast, the sum vector points from her home to where she ends up. (p. 551–552)

Scaling vectors. Multiplying by a constant changes the length by that factor and keeps the direction. (pp. 552–553)

$3\vec u$ is three times as long, in the same direction. It is the same as $\vec u+\vec u+\vec u$ .
$-\vec u$ has the same length but points the opposite way. Adding it to $\vec u$ brings you back to the start.
$-2\vec u$ is twice as long as $\vec u$ and points the opposite way.

Subtracting vectors. Subtraction is addition of the opposite: $\vec u-\vec v=\vec u+(-\vec v)$ . To draw it, start with $\vec u$ , draw $-\vec v$ from its end, and join the start of $\vec u$ to the end of that arrow. If $\vec u$ and $\vec v$ are drawn from the same point as two edges of a parallelogram, then $\vec u+\vec v$ and $\vec u-\vec v$ are the two diagonals of that parallelogram. (pp. 553–554)

Describing a Vector with Numbers

Pictures give a feel for vectors, but calculations need numbers. There are two main ways to describe a vector numerically.

Magnitude and direction. The length is written $\|\vec u\|$ , and the direction is an angle $\theta$ , usually measured from standard position (from the positive x-axis, counterclockwise). A vector of length 1 is a unit vector. Dividing any vector by its length produces a unit vector. (p. 554)

Component form. The vector $\langle x,y\rangle$ means a horizontal displacement of $x$ and a vertical displacement of $y$ . The same vector can be written $x\vec i+y\vec j$ , where $\vec i=\langle1,0\rangle$ points right and $\vec j=\langle0,1\rangle$ points up. So $\langle3,-4\rangle=3\vec i-4\vec j$ means 3 units right and 4 units down. A vector shows a change in position, not a fixed location, so it can be drawn starting anywhere. (pp. 554–555)

The magnitude is the hypotenuse of the right triangle with legs $x$ and $y$ . This is the same as the radius $r$ in polar form.

Converting Between the Two Forms

The conversions are the same as polar to Cartesian coordinates. (p. 555)

From magnitude and angle to components: $x=r\cos\theta,\qquad y=r\sin\theta$ Example: a vector of length 7 at $135^\circ$ has $x=7\cos135^\circ=-\tfrac{7\sqrt2}{2}$ and $y=7\sin135^\circ=\tfrac{7\sqrt2}{2}$ . So it is $\left\langle-\tfrac{7\sqrt2}{2},\tfrac{7\sqrt2}{2}\right\rangle$ .

From components to magnitude and angle: $r=\sqrt{x^2+y^2},\qquad \tan\theta=\frac{y}{x}$ Example: for $\langle3,-2\rangle$ , $r=\sqrt{9+4}=\sqrt{13}$ . Then $\tan^{-1}(-2/3)\approx-33.69^\circ$ , which equals $326.31^\circ$ . This is in the 4th quadrant, which matches $x>0,\ y<0$ . The inverse tangent only returns angles between $-90^\circ$ and $90^\circ$ , so always check the signs of $x$ and $y$ . If the angle lands in the wrong quadrant, add $180^\circ$ . (pp. 555–556)

Velocity and force. Any quantity with size and direction can be treated as a vector, including velocity and force. Splitting it into horizontal and vertical parts is called resolving it into components. An object launched at 200 m/s at $30^\circ$ has horizontal velocity $200\cos30^\circ\approx173$ m/s and vertical velocity $200\sin30^\circ=100$ m/s. (p. 556)

Arithmetic in Component Form

In component form, you add, subtract, and scale one component at a time. (p. 557) $\langle u_1,u_2\rangle\pm\langle v_1,v_2\rangle=\langle u_1\pm v_1,\;u_2\pm v_2\rangle,\qquad c\langle u_1,u_2\rangle=\langle cu_1,cu_2\rangle$

Example: let $\vec u=\langle3,-2\rangle$ and $\vec v=\langle-1,4\rangle$ . Scale first: $3\vec u=\langle9,-6\rangle$ and $2\vec v=\langle-2,8\rangle$ . Then subtract: $3\vec u-2\vec v=\langle9-(-2),\,-6-8\rangle=\langle11,-14\rangle$ .

Combining Several Movements

When there are three or more movements, drawing triangles gets messy. A better method has three steps. (p. 557)

Write each movement in component form.
Add the x-components and add the y-components.
If the question needs it, convert the result to magnitude and direction.

Compass directions. Maps put north up and east to the right. So NE, SE, SW, and NW are $45^\circ$ diagonals. If you use the standard-position angle for each movement, the cosine and sine give the correct signs automatically. Southeast, for instance, is $315^\circ$ . (pp. 557–558)

Example. A woman walks 3 miles north, then 2 miles southeast.

North: $\langle0,3\rangle$ .
Southeast: $\langle2\cos315^\circ,2\sin315^\circ\rangle\approx\langle1.414,-1.414\rangle$ .
Sum: $\langle1.414,1.586\rangle$ .
Distance from home: $\sqrt{1.414^2+1.586^2}\approx2.125$ miles.
Direction away from home: $\tan^{-1}(1.586/1.414)\approx48.27^\circ$ north of east. To return she heads the opposite way, at $180^\circ+48.27^\circ=228.27^\circ$ .

Common mistake: displacement is not distance traveled. The displacement vector is the straight-line arrow from start to end, about 2.125 miles here. The total distance walked adds every leg, including the walk home: $3+2+2.125=7.125$ miles. These two quantities are different. (p. 558)

Forces in Equilibrium

An object is in equilibrium when all the forces on it add to $\langle0,0\rangle$ . To find the missing force, resolve each known force into components and add them. The needed force is the negative of that sum. (pp. 559–560)

Example: three forces act on an object: 6 N at $30^\circ$ , 7 N pointing left, and 4 N at $300^\circ$ .

$6\langle\cos30^\circ,\sin30^\circ\rangle=\langle3\sqrt3,3\rangle$
$\langle-7,0\rangle$
$4\langle\cos300^\circ,\sin300^\circ\rangle=\langle2,-2\sqrt3\rangle$

The sum is $\langle3\sqrt3-5,\;3-2\sqrt3\rangle\approx\langle0.196,-0.464\rangle$ . The balancing force is the opposite, $\langle-0.196,0.464\rangle$ . Its magnitude is $\sqrt{(-0.196)^2+0.464^2}\approx0.504$ N. The calculator gives $\tan^{-1}(0.464/(-0.196))\approx-67.09^\circ$ . This is in the wrong quadrant, because $x<0$ and $y>0$ place the vector in the 2nd quadrant. Adding $180^\circ$ gives $112.91^\circ$ . So the extra force is about 0.504 N at $112.9^\circ$ .

The dot product and the angle between vectors

The dot product turns two vectors into one number. There are two ways to compute it, and they always agree.

Component form: $\vec a\cdot\vec b=a_1b_1+a_2b_2$ . Multiply the x-parts, multiply the y-parts, and add.
Geometric form: $\vec a\cdot\vec b=\|\vec a\|\|\vec b\|\cos\theta$ , where $\theta$ is the angle between the vectors.

Use the component form when you have coordinates. Use the geometric form when you know lengths and an angle. Example: $\langle3,5\rangle\cdot\langle2,-1\rangle=6-5=1$ . (pp. 565–566)

Setting the two forms equal and solving for the angle gives

$\theta=\cos^{-1}\!\left(\frac{\vec a\cdot\vec b}{\|\vec a\|\|\vec b\|}\right).$

Steps: find the dot product, find both magnitudes, divide, then take the inverse cosine. For $\langle5,1\rangle$ and $\langle4,2\rangle$ the dot product is $20+2=22$ . The magnitudes are $\sqrt{26}$ and $\sqrt{20}$ . So $\theta=\cos^{-1}\!\left(\frac{22}{\sqrt{26}\sqrt{20}}\right)\approx15.255^\circ$ . (p. 567)

The sign of the dot product tells you about the angle without any inverse cosine, because the dot product has the same sign as $\cos\theta$ :

Zero: the angle is $90^\circ$ , so the vectors are orthogonal (perpendicular).
Positive: the angle is less than $90^\circ$ .
Negative: the angle is greater than $90^\circ$ .

Example: $\langle6,4\rangle\cdot\langle-2,3\rangle=-12+12=0$ , so these vectors are perpendicular. (pp. 567–568)

Projections and orthogonal components

A projection answers the question: how much of $\vec a$ points along $\vec b$ ? Picture a right triangle. $\vec a$ is the hypotenuse, and the projection lies along $\vec b$ as the adjacent side. Its length is $\|\vec a\|\cos\theta$ . Replacing $\cos\theta$ with the dot-product expression gives a formula that needs no angle:

$\mathrm{comp}_{\vec b}\vec a=\frac{\vec a\cdot\vec b}{\|\vec b\|}.$

This is only a length. To get the projection as a vector, give that length the direction of $\vec b$ . Divide $\vec b$ by $\|\vec b\|$ to get a unit vector, then multiply:

$\mathrm{proj}_{\vec b}\vec a=\frac{\vec a\cdot\vec b}{\|\vec b\|^2}\,\vec b.$

Example: project $\vec a=\langle3,-2\rangle$ onto $\vec b=\langle8,6\rangle$ .

$\vec a\cdot\vec b=24-12=12$ .
$\|\vec b\|=10$ , so the length is $12/10=6/5$ .
The vector is $\frac{12}{100}\langle8,6\rangle=\langle\tfrac{24}{25},\tfrac{18}{25}\rangle$ . (pp. 568–569)

Orthogonal component. Any vector $\vec a$ splits into two pieces: the projection $\vec u$ along $\vec b$ , and a leftover piece $\vec v$ at a right angle to $\vec b$ . Since $\vec a=\vec u+\vec v$ , we get $\vec v=\vec a-\mathrm{proj}_{\vec b}\vec a$ . Example: for $\vec a=\langle-3,4\rangle$ and $\vec b=\langle-8,4\rangle$ , the projection is $\langle-4,2\rangle$ . So $\vec v=\langle-3,4\rangle-\langle-4,2\rangle=\langle1,2\rangle$ . Check: $\langle1,2\rangle\cdot\langle-8,4\rangle=0$ . (p. 570)

Work as a dot product

When a constant force moves an object, only the part of the force along the motion does work. That part is the projection of $\vec F$ onto the displacement $\vec d$ . Multiplying by the distance gives

$W=\vec F\cdot\vec d=\|\vec F\|\|\vec d\|\cos\theta.$

Example: a 30 lb pull on a rope at $30^\circ$ drags a cart 20 ft. Then $W=30\cdot20\cdot\cos30^\circ\approx519.6$ ft-lb. If you are given points instead, first subtract them to get the displacement vector, then use components. (p. 570)

Parametric equations

A system of parametric equations gives $x(t)$ and $y(t)$ , both in terms of a third variable $t$ , the parameter. Often $t$ is time. This lets you describe curves that fail the vertical line test, such as circles, and also describe how something moves along a path. Example: $x(t)=2t-5$ , $y(t)=3-t$ gives an object's position at time $t$ . (p. 574)

Steady straight-line motion. Find how fast each coordinate changes, then write each as a linear function of $t$ . Moving from $(-5,3)$ to $(3,-1)$ in 4 seconds, x changes by 8 (2 per second) and y by $-4$ ( $-1$ per second). So $x(t)=-5+2t$ and $y(t)=3-t$ . A table of $t$ , $x$ and $y$ values lets you plot the path, with arrows showing the direction of motion. (pp. 574–575)

Many representations. The same path can have different parameterizations. Using $x(t)=-5+2t^2$ and $y(t)=3-t^2$ gives the same segment between the same two points. But the object now arrives at $t=2$ , not $t=4$ . Changing how $t$ enters changes the speed, not necessarily the path. (p. 575)

Circles, ellipses and other standard curves

$x=r\cos t$ , $y=r\sin t$ traces a circle of radius $r$ counterclockwise, starting at $(r,0)$ .
$x=r\sin t$ , $y=r\cos t$ traces the same circle clockwise, starting at $(0,r)$ .
$x=a\cos t$ , $y=b\sin t$ traces an ellipse. For $x=2\cos t$ , $y=3\sin t$ , x runs from $-2$ to $2$ and y from $-3$ to $3$ .

Graphing calculators need parametric mode, where you enter an $x(t)=$ and $y(t)=$ pair. (pp. 576–577)

Converting between parametric and Cartesian form

Parametric to Cartesian. Solve one equation for $t$ and substitute into the other. For $x=t^2+1$ , $y=t+2$ : $t=y-2$ , so $x=(y-2)^2+1$ . This is a parabola with vertex $(1,2)$ opening right. (pp. 578–579)

The range of $t$ can limit the Cartesian domain. For $x=\sqrt t+2$ , $y=\log t$ with $t>0$ , we get $y=\log\big((x-2)^2\big)$ , but only for $x>2$ .

For trig equations, avoid inverse trig functions and use an identity instead. From $x=2\cos t$ and $y=3\sin t$ , put $\cos t=x/2$ and $\sin t=y/3$ into $\cos^2t+\sin^2t=1$ . This gives $\frac{x^2}{4}+\frac{y^2}{9}=1$ . (pp. 579–580)

Cartesian to parametric. If one variable is a function of the other, let the independent variable equal $t$ . For $x=y^3-2y$ , set $y=t$ , so $x=t^3-2t$ . For a circle such as $x^2+y^2=3$ , use sine and cosine: $x=\sqrt3\cos t$ , $y=\sqrt3\sin t$ . (pp. 580–581)

Rolling wheels and combined circular motion

Cycloid. A light on the rim of a wheel of radius $r$ rolls forward. After the wheel turns through angle $\theta$ , the center is at $(r\theta,\,r)$ . The arc length $s=r\theta$ is the distance rolled, and the center stays at height $r$ . The light sits at $(-r\sin\theta,\,-r\cos\theta)$ relative to the center. Add the two positions:

$x=r\theta-r\sin\theta,\qquad y=r-r\cos\theta.$

(pp. 581–582)

Motion within motion. If one circular motion is carried along another, add the x equations and add the y equations. A circular motion with period $P$ has angular rate $2\pi/P$ . A moon orbits a planet with radius 6 and period 10 days, and the planet orbits a sun with radius 30 and period 100 days. The moon's position relative to the sun is

$x=6\cos\!\left(\tfrac{\pi}{5}t\right)+30\cos\!\left(\tfrac{\pi}{50}t\right),\quad y=6\sin\!\left(\tfrac{\pi}{5}t\right)+30\sin\!\left(\tfrac{\pi}{50}t\right).$

(pp. 582–583)

Check yourself

Two sides and a non-included angle of a triangle are known. Why can this give 0, 1 or 2 triangles, and what does a sine value above 1 tell you?
Convert the polar point $(4,\tfrac{2\pi}{3})$ to Cartesian coordinates, and give one other polar representation of it.
Use DeMoivre's Theorem to write $(1+i)^4$ in polar form and then in the form $a+bi$ .
Find the angle between $\langle2,1\rangle$ and $\langle-1,3\rangle$ , and find the projection of $\langle2,1\rangle$ onto $\langle-1,3\rangle$ .
Eliminate the parameter from $x=3\cos t$ , $y=3\sin t$ to get a Cartesian equation, and state the direction of motion.

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