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 between the beam and the line through the searchlight perpendicular to the wall is pi/4? Note that dTheta/dt=3(2pi)=6pi.
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b= 15; tan(t) should give us a way to relate the light on the wall to the distance to the search light
15 miles is pretty far
tan(t) = wb^-1 sec^2(t) t' = w'b^-1 w' = sec^2(t) t'b
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without having to guess at what you fail to grasp and try to teach you what you should have already been exposed to, why dont you try to express your confusion in a manner that can be communicated?
I apologize. I suffer from a learning disability and it takes a lot of work for me to understand this. I suppose my problem is structuring this in a way I understand.
ok, these rates of change problems require us to relate what we know to what we are trying to determine. in this case we need to relate the speed of the angle with the speed of the spotlight against the wall the triangle i drew shows us all the parts and we can relate them together with the tangent function.
by taking the derivative of that relationship we construct an equation tat relates speeds together. tan(t) = w/b sec^2(t) t' = w'/b we want to know w' so rearrange the eq to solve for w' w' = b sec^2(t) t'
b = 15, t = pi/4, and t' = 6pi ; as given
soo w' = 15 sec^2(pi/4) *6pi
w'=15*2*6pi if i see it right
which is 565.49?
180pi is all i can tell without a calculator
it wants it in mph
and what we found is in miles per minute aint it
60 minutes in an hour; so multiply by 60 to get hours
multiply 180pi * 60 ?
yeah, thats what it looks like to me. with any luck im reading it right
10,800?
yep
ill have to dbl chk my work to be sure tho
well that was wrong. crap.
do you have to keep it in pi or estimate it?
doesnt say.
you sure you got all the info correct in the post? 15 miles, pi/4 and 6pi per minute?
at 3 revolutions a minute yes
15 miles seems big to me; i want to see it as 1.5
its fifteen
then 10,800 pi is the best i can come up with :)
I hate webworks
yeah, i loathe having to work with a program
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