Couple of Math Questions [Precalculus]?

Question

1) Suppose t is the length of arc on the unit circle whose initial point (1,0) , and terminal point is P(x,y). If t = 4.9, sketch t on the unit circle for the approximate location of P(x,y). 
[If someone could just tell me which quadrant it would be in, it would help]

2) The following function with equation f(x) = 5 sin (2x-pi/6) when graphed will have the following characteristics:
a) Amplitude: 5?
b) Phase shift: pi/12?
c) Length of one period: ?

3) The following function with equation f(x) = ln (x+3) +2 has a vertical asmptote and an x-intercept.
a) Equation of vertical asymptote: x=?
b) x-intercept: (4-decimal place accuracy)

Astral Walker · Accepted Answer

1) You generally presume that you travel counterclockwise around the unit circle.

Since we are dealing with the unit circle, we know the total arclength of the circle is 2pi (the circumference). So to determine the relative position of t you divide 4.9 by 2pi and get

4.9/(2pi) = 0.78

0-0.25 is the 1st quadrant
0.25-0.5 is the 2nd quadrant
0.5-0.75 is the 3rd quadrant
0.75-1 is the 4th quadrant

so t is just entering in the 4th quadrant so t should be roughly a bit to the right of the y axis near the point (0,-1) and should be around (a,b) where a is a bit larger than 0 and b is a bit smaller than -1.

2a) yes amplitude is the coefficient of the sine function

2b) Since the frequency is 2 (x from 0 to pi completes 1 cycle) then the shift of pi/6 is over 2 cycles so it is pi/12 for 1 cycle so this appears correct

2c) the length of 1 period is pi since x from 0 to pi will map 2x from 0 to 2pi.  Therefore there are 2 waves per cycle so the frquency is 2 and the wavelength is 1/2 cycle

3a) A vertical asymptote occurs when lim f'(x) -> infinity

f'(x) = 1/(x+3) so when x=-3 there is a vertical asymptote.

3b) The y intercept occurs at f(0) but the x intercept occurs when f(x) = 0 so solve  f(x) for f(x)=0

f(x) = ln(x+3) + 2

ln(x+3) + 2 = 0

ln(x+3) = -2

Now take the exponential of both sides

e^ln(x+3) = e^-2

x+3 = e^-2

x = e^-2 - 3

x = 0.13534 - 3 = -2.86466 ~ -2.8647

davidosterberg1 · Answer

circumference of a unit circle = 2TT = 6.28.
4.9 / 6.28 = .78. From (1,0) travel 78% of the distance counterclockwise, and end up at P. [approx 270degrees]  (0,-1)

amplitude of 5

someone else · Answer

use angular equation 
theta times radius=arc length
theta times 1=4.9
theta=4.9radian=280degree
280 degree from (1,0) is 1st or 4th quadrant

length of period is pi

equation of vertical asumptote is x=-3
x intercept is -2.8647

? · Answer

cos x tan x - sin^2 x = 0 cos x • sin x/cos x - sin^2 x = 0 sin x - sin^2 x = 0 sin x(a million - sin x) = 0 sin x = 0 or a million - sin x = 0 x = 0, ?, .... a million = sin x x = 2k?........x = ?/2, 5?/2, ... ....................x = (4k + a million) ?/2 +++++++++++++++++++++ 2 sin^2 x - sin x = 0 sin x(2 sin x - a million) = 0 sin x = 0 or 2 sin x - a million = 0 x = 0, ?............2 sin x = a million ............................sin x = a million/2 ...........................x = ?/6, 5?/6

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Couple of Math Questions [Precalculus]?

4 Answers