LaddiusMaximus
  • LaddiusMaximus
A searchlight rotates at a rate of 3 revolutions per minute. The beam hits a wall located 15 miles away and produces a dot of light that moves horizontally along the wall. How fast (in miles per hour) is this dot moving when the angle \theta between the beam and the line through the searchlight perpendicular to the wall is pi/4? Note that dtheta/dt3(2pi)=6pi
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The horizontal distance is 11 miles * tan(pi/6) = 6.35 miles [remember pi/6 is radians not degrees] At 6pi rad/min, the angle pi/6 is swept out in pi/6 /(6pi) min = 1/36 min = 1.67 sec So the dot speed is 6.35 miles per 1.67 sec = 3.8 miles / sec = 228 miles per minute = 13,690 mph
LaddiusMaximus
  • LaddiusMaximus
thats not right
amistre64
  • amistre64
ive been thinking about this one for a few days :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
is the 15 miles a perpendicular distance from the wall?
LaddiusMaximus
  • LaddiusMaximus
yes
amistre64
  • amistre64
|dw:1332800881728:dw|
amistre64
  • amistre64
|dw:1332801016388:dw|ok, one thing I thought of was relating the linear speed of the circumference of the circle created by the beam 15sqrt(2) * 2 pi = 30sqrt(2)pi, at 3 times a minute = 90sqrt(2) pi this isnt the speed along the wall, this is the speed along the circle that hits the wall at that point
LaddiusMaximus
  • LaddiusMaximus
all this has me so confused
amistre64
  • amistre64
im wondering if the rotation speed can be component based to measure the wall speed. |dw:1332801109026:dw| which would make the linear speed to be 90pi miles per minute. if its to be in hours we would then times that by 60 5400pi mph IF thats a workable method. which I aint all that sure about :)
LaddiusMaximus
  • LaddiusMaximus
yeah thats not it either. sheesh
amistre64
  • amistre64
it would help if you could gives us reasons as to why not ... other than simply becasue the program your inputing it into says its wrong :/
LaddiusMaximus
  • LaddiusMaximus
Yeah me too. But webworks doesnt give a reason. It just tells me its wrong. Helpful, right??
amistre64
  • amistre64
yeah, there aint a "how to do a similar" problem option is there?
LaddiusMaximus
  • LaddiusMaximus
nope. The calc department couldnt have found a worse program to use.
LaddiusMaximus
  • LaddiusMaximus
They reference a problem in the book but thats about it.
phi
  • phi
I think it's reasonable to use the instantaneous speed r dtheta when r= 15 sqrt(2) dtheta= 6pi rad/1min or 360pi rad / hour so r dtheta = 15*360*sqrt(2)*pi in mph about 23991 mph
LaddiusMaximus
  • LaddiusMaximus
yeah that one isnt right either.
phi
  • phi
'course I could be wrong.
amistre64
  • amistre64
i think thats abt what i had last time tan(t) = w/15 15sec^2(t) t' = w' 15*2*t' = w' 30*t' = w'
amistre64
  • amistre64
180pi*60 = 10800pi mph was last time
LaddiusMaximus
  • LaddiusMaximus
maybe the program is wrong
amistre64
  • amistre64
is it spose to be exact? or estimated with 3.14?
LaddiusMaximus
  • LaddiusMaximus
doesnt say
phi
  • phi
How about starting with the wall modeled as a vertical line in polar coords r cos(A)= 15 and take the derivative
amistre64
  • amistre64
http://wp-blogs.moundsparkacademy.org/tro/files/2011/08/chap03.11-et-student-solutions1.pdf
amistre64
  • amistre64
#27, same method I used the first time around
amistre64
  • amistre64
15*6pi*(sqrt(2))^2 = 180pi * 60 = 10800pi mph
amistre64
  • amistre64
ironically, thats twice the speed i thought of with the circle strategy :)
LaddiusMaximus
  • LaddiusMaximus
then the program is wrong
LaddiusMaximus
  • LaddiusMaximus
ok got it. forgot pi
amistre64
  • amistre64
..... i think i just had a synapse burst :/
phi
  • phi
So what is the final answer??
LaddiusMaximus
  • LaddiusMaximus
10800pi

Looking for something else?

Not the answer you are looking for? Search for more explanations.