Check my answers please! This is the second part of my final and I just want to make sure I did everything correctly.
1. Graph the function f(x) = -sec(x/4)+3 on the interval [-2pi,6pi].
I'm having an awful time with this one. I'm able to graph the basic equation, but I'm confused on how to incorporate the interval into it. Does it actually change the graph in any way? How would I do this one?
2. Evaluate (5 + 5sqrt3i)^7 using DeMoivre's theorem. Write our your answer in rectangular form.
r = sqrt(a^2 + b^2) and theta = tan^-1(b/a)
R = sqrt (5)^2 + (5sqrt3)^2
R = sqrt 25 + 75
So r = 10
Theta = tan^-1 (b/a)
T = tan^-1 ((5sqrt3)/5)
T = pi/3 or 60 degrees
So (5 + 5sqrt3i)^7 is equal to 10cis(60degrees)
= 10^7 cis(60)
X = rcostheta
Y = rsintheta
X = 10^7 cos(60) = 10^7 * ½ = 5 * 10^6
Y = 10^7 sin (60) = 10^7 * sqrt3/2 = 5sqrt3 * 10^6
So for our final answer: 10^7(1 + I sqrt3)/2
3. A helicopter departs from an airport flying at a speed of 90 miles per hour at a bearing of 250° for 20 minutes before landing on a helipad. Determine the helipad’s location in relation to the airport. Round to the nearest tenth of a mile. Show all your work and explain your reasoning.
1 hr = 60 min
90/60 = 1.5 miles/min
1.5 x 20 = 30
So the distance the helicopter traveled is 30 miles.
30 cos 250 degrees = -10.3
30 sin 250 degrees = -28.2
The helipad would be about 28.2 miles West and 10.3 miles South of the airport.
Very thoughtful post, guest! 2&3 looks good to me but I do not know what sec is.
1. Graph the function f(x) = -sec(x/4)+3 on the interval [-2pi,6pi]
The " -" flips the normal secant graph over the x axis
The (x/4) lengthens the normal period by a factor of 4
The " +3 " shifts everything up by 3 units
Here's the normal graph and the transformed graph on [-2pi, 6pi]