Download Thursday Apr 10 - UGA Math Department

Test 5 Review
(Note: there’s less of a summary here than in previous tests to save space. You should make your own
formula sheets for practice.)
A few summary remarks:
• Break up complex diagrams into simple shapes. Do you get right triangles (SOHCAHTOA), or do you
find SAS Area?
– Other useful tools: a2 + b2 + c2 (right triangles with two known sides), and α + β + γ = 180◦ (when
you know two angles)
– Take advantage of symmetry. For instance, with a regular polygon, how many triangles do you break
it into? What’s the central angle in each triangle? Are these right triangles or SAS?
– Sometimes it’s easier to subtract parts to find your answers. (See the “water slide” problem.)
– Elevation angle = depression angle (alternate interiors). Measure these against a horizontal axis!
• Sometimes it’s much easier to use reciprocal trig functions when writing answers, to avoid having as many
fractions. This happens especially when you have your unknown in a denominator.
SAMPLE:
tan(15deg) =
100
x
⇒
x=
100
= 100 cot(15deg)
tan(15deg)
Selected WebQuiz 9 and 10 Problems with Hints
For an overview of more problems, visit my website and look for an extra downloadable handout. Don’t
forget to also look for my extra attachments for this unit!
NOTE: The end of WQ 9 and beginning of WQ 10 are quite challenging. WQ 9 needs good picture-labeling
skills. WQ 10 needs more algebraic care with solving trig equations.
WebQuiz 9:
#5: “Suppose f (t) measures the temperature in ◦ C at t hours, where t = 0 means midnight. Assume f (t)
is decreasing at midnight and the function f (t) = a sin(bt + c) + d has period 24. Say the high temperature of
20◦ C occurs at 2 P.M., and the average temperature of 12◦ C occurs 6 hours later. If a, b > 0 and c is in [−π, π],
then find the values of a, b, c, d.”
This question is a lot like the problems that give you a picture and ask you to figure out its amplitude, period,
and phase shift first. We aren’t given the high and low temp: we’re given high and AVERAGE. However, you
can still get a and d from this. You are told what the period is, so you can get b.
The phase shift is tricky. Recall that sine starts on the axis and rises. In other words, it occurs BEFORE
the maximum, not after! Instead of using the time 6 hours later than 2pm for phase shift, what should we use?
(Also think: 6 hours is a quarter-period here!)
#6: “A function f (x) has the form f (x) = a + sin(bx) where a and b are constants, and −7/2 < b < 7/2.
If f (0) = 4 and f (π/7) = 3.5, find a and b.”
We’ve seen problems like these when we covered 5.5 and 5.6. You can use the intercept (0, 4) to get a.
Plugging in the point (π/7, 3.5), you should get sin(bπ/7) = −1/2. We should treat this like a two-layer
problem with θ = bπ/7 on the inside: you should get an interval of −π/2 to π/2 for θ, and you should find
θ needs to be in Quadrant IV. However, this is one of the cases where Quadrant IV uses negative angles, so
θ = −θR here! (Draw Quadrants I and IV and note the symmetry...)
#8: “A circle with radius 10 inches is inscribed in
If you change the number of sides in the polygon,
a regular hexagon. OA is a radius of the circle. Find where does that change affect the work? Which steps
the area of the region between the hexagon and the cir- stay the same?
cle, then generalize this approach to find the area that
would be between a 15-sided polygon and the circle.”
This is one of the problem types we’ve seen with regular polygons and a circle. Notice the radius touches
the middle of the sides: that’s why the triangles we
use to break up the hexagon taken up only half a side!
That means we have 12 of those triangles making up
the hexagon, instead of six. You can find the central
angle in each triangle, and use that with the adjacent
side of 10 (the height of the triangle) to get the base
by SOHCAHTOA.
#9: “A ladder 16 feet long leans against the side of a building. The angle between the ladder and the
building is 32◦ .”
(a) Approximate the distance from the bottom of the ladder to the building.
(b) If the distance from the bottom of the ladder to the building increases by 1.5 feet, how far does the top
of the ladder move down the building?
The first part of this question is fairly standard work with SOHCAHTOA, but be careful about labeling
your triangle. The ladder lies diagonally, so the hypotenuse is 16. The angle given is made with the building,
not the ground, so you label the angle at the top of the triangle!
For the second part, the base of the triangle has now changed. Thanks to part (a), you knew the old base’s
length, so now you know the new base length. Unfortunately, the triangle’s angles have changed now, so you
can’t use SOHCAHTOA, but you do know two sides of the triangle. If you figure out the third side, which is
the new height of the triangle, how does that help you finish the problem?
#10: “The drawbridge in the picture is l = 100 feet long, and its sections can open up to α = 32◦ . The
water level is 15 feet below the closed bridge. When the bridge is fully opened, find (a) the height d of the
bridge ends over the water, and (b) the distance between the bridge ends.”
This is quite a lot like the water slide problem from
HW 5.7. Each section of the bridge makes a right triangle where the bridge forms the hypotenuse and α is
the elevation angle. Since there are two triangles, the
bridge is split in two, so each hypotenuse is l/2.
One part of the question requires you to find the
height of this triangle, and the other part requires you
to find the base. Which is which, and why?
WebQuiz 10:
#3: “What is the domain of the function f (x) = log(sin x)? (Pick one choice.)”
• (0, ∞)
• (0, π) only
• (−π/2, π/2) only
• (2kπ, 2kπ + π) for integers k
• (2kπ − π/2, 2kπ + π/2) for integers k
Recall that logarithms are NOT always defined: the inside of the logarithm needs to be positive. Thus,
we want sin x > 0. When you do tricky inequalities, a number line is your best tool! Find the places where
EQUALITY occurs, and then check the intervals you get.
You can show that sin x = 0 when x = πn for any integer n, so our number line breaks up into intervals of
length π. By checking some values in the middle like π/2 or 3π/2, you’ll see that (0, π) should be part of the
answer (as sin is positive on that interval) whereas (π, 2π) should not be. Given that we have a period of 2π,
which OTHER intervals should be part of the answer?
#4: “Find the exact solutions of 3 sin x − 3 csc x = 0 for x in [0, 2π).”
Remember that csc = 1/ sin, so you can multiply sin throughout the equation to get rid of the fractions!
This gives you a quadratic in sin x.
#4 variant: “Solve sec4 (x) = 16/9 for all x in [0, 2π).”
I would take reciprocals to get cos4 (x) = 9/16. There are two fourth-roots (positive
negative), and it’s
p√ pand
√
√
easiest to simplify them if you think of taking square roots twice. You get ±
9/
16, or ± 3/2. You
should get answers in all four quadrants (why?).
#10: “A tidal wave of height 40 feet has a period of 18 minutes. The distance y from sea level to the top of
the wave is y = 20 cos(π/9 · t) with t in minutes. In each 18-minute period, how many minutes does the wave
spend BELOW a wall whose height is 10 feet above sea level?”
This one is hard! When you cut through all the wording, though, the question is really asking you to figure
out when y < 10, which solves to cos(π/9 · t) < 1/2. This actually combines two tricks from previous problems
in these quizzes!
1. First, we want to find the places where cos(π/9 · t) DOES equal 1/2, where 0 ≤ t ≤ 18 (that’s one period).
Once we have those, we can mark intervals and figure out which ones are relevant to our answer. (This is
what we did in #3 on this quiz.)
2. Say θ = π/9t, and you can check 0 ≤ θ ≤ 2π (i.e. one period of angles... this comes from #6 in WQ 9).
The two solutions to cos(θ) = 1/2 are π/3 and 2π/3, so θ = π/9t = π/3, 2π/3. This gives you t = 3, 6.
Now, test some values of t and ask yourself which times put the wave below the wall: is it the 3 minutes
between t = 3 and t = 6, or it is the remaining 15 minutes of the period outside of that interval?