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Mathematics 22 Online
OpenStudy (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 between the beam and the line through the searchlight perpendicular to the wall is pi/4? Note that dTheta/dt=3(2pi)=6pi.

OpenStudy (amistre64):

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OpenStudy (amistre64):

b= 15; tan(t) should give us a way to relate the light on the wall to the distance to the search light

OpenStudy (amistre64):

15 miles is pretty far

OpenStudy (amistre64):

tan(t) = wb^-1 sec^2(t) t' = w'b^-1 w' = sec^2(t) t'b

OpenStudy (laddiusmaximus):

?

OpenStudy (amistre64):

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?

OpenStudy (laddiusmaximus):

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.

OpenStudy (amistre64):

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.

OpenStudy (amistre64):

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'

OpenStudy (amistre64):

b = 15, t = pi/4, and t' = 6pi ; as given

OpenStudy (amistre64):

soo w' = 15 sec^2(pi/4) *6pi

OpenStudy (amistre64):

w'=15*2*6pi if i see it right

OpenStudy (laddiusmaximus):

which is 565.49?

OpenStudy (amistre64):

180pi is all i can tell without a calculator

OpenStudy (laddiusmaximus):

it wants it in mph

OpenStudy (amistre64):

and what we found is in miles per minute aint it

OpenStudy (amistre64):

60 minutes in an hour; so multiply by 60 to get hours

OpenStudy (laddiusmaximus):

multiply 180pi * 60 ?

OpenStudy (amistre64):

yeah, thats what it looks like to me. with any luck im reading it right

OpenStudy (laddiusmaximus):

10,800?

OpenStudy (amistre64):

yep

OpenStudy (amistre64):

ill have to dbl chk my work to be sure tho

OpenStudy (laddiusmaximus):

well that was wrong. crap.

OpenStudy (amistre64):

do you have to keep it in pi or estimate it?

OpenStudy (laddiusmaximus):

doesnt say.

OpenStudy (amistre64):

you sure you got all the info correct in the post? 15 miles, pi/4 and 6pi per minute?

OpenStudy (laddiusmaximus):

at 3 revolutions a minute yes

OpenStudy (amistre64):

15 miles seems big to me; i want to see it as 1.5

OpenStudy (laddiusmaximus):

its fifteen

OpenStudy (amistre64):

then 10,800 pi is the best i can come up with :)

OpenStudy (laddiusmaximus):

I hate webworks

OpenStudy (amistre64):

yeah, i loathe having to work with a program

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