A laser rangefinder is locked on a comet approaching earth...

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 42 days, is given by g(x)=200,000csc(π42x). b. Evaluate g(7). Enter the exact answer. g(7)= c. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? The minimum distance between the comet and Earth is km which is the . It occurs at x days. d. Find and discuss the meaning of any vertical asymptotes on the interval [0,49]. The field below accepts a list of numbers or formulas separated by semicolons (e.g. 2;4;6 or x+1;x−1 ). The order of the list does not matter. x= At the vertical asymptotes the comet is out of range or no longer in orbit?

Transcript

00:01 In this problem, we've been given this co -sequent function, 200 ,000 co -seacant of pi over 42 times x.

00:08 We know that we're looking at the interval of x from 0 to 42, and we are told that this function g of x is giving us the distance between earth and an approaching comet over these 42 days.

00:24 So when we find g of 7, what we're doing is we're finding the distance between the earth and the comet, after seven days.

00:33 So let's plug this in and let's see what we can get.

00:35 We have 200 ,000 times the co -secant of pi over 42 times seven.

00:48 So this is going to end up being two times the co -secant of seven pi over 42, which reduces to pi over six.

01:10 All right, so we need to go back to our unit circle and we need to think about how co -seant is related to the other trick functions that we do know.

01:19 We know that co -sequent is the reciprocal of sign.

01:22 So this is really 200 ,000 times 1 over the sign of pi over 6.

01:29 And we know that pi over 6, if you think in degrees, is 30 degrees.

01:35 P pi over 6 is 30 degrees.

01:37 Sign is the x value on the unit circle at pi over 6.

01:41 So that's going to come out to be one half.

01:45 So this is 200 ,000 times 1 over 1 half, which is just 200 ,000 times 2.

01:58 So this comes out to be 400 ,000.

02:00 So the distance is 400 ,000 kilometers at seven days.

02:07 Then we want to know what is the minimum distance, what is the closest this gets to earth? and for this, we need to remember, let's remember what cosecican looks like.

02:17 Co -secant is going to be the reciprocal of sign.

02:20 So if we have sign at looking like this, it's going to be a value of 1 and a value of negative 1 here, and it oscillates between this, i missed my point...

Question

Please give Ace some feedback