trig. and rates of change (1 Viewer)

[freaking_out]

1. The angle x between two radii OP and OQ of a circle of radius 6cm is increasing at the rate of 0.1 radians per minute. Find the value of x for which the rate of increase of the area of the segment cut off by the chord PQ is at its maximum.

2. A rotating light L is situated at sea 180 metres form the nearest point P on a straight shoreline. The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136pi m/s.


thanx in advance

 

[kini mini]

Originally posted by freaking_out
1. The angle x between two radii OP and OQ of a circle of radius 6cm is increasing at the rate of 0.1 radians per minute. Find the value of x for which the rate of increase of the area of the segment cut off by the chord PQ is at its maximum.

I have temporarily fogotten which bit the segment is

2. A rotating light L is situated at sea 180 metres form the nearest point P on a straight shoreline. The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136pi m/s.

An eerily familiar question! First, draw a diagram.

Let M be any point on the shoreline distinct from P.

Let angle MPL = x
Let length PM = l

Then tan x = l/180

l = 180*tan x

since the light rotates once evry 10 seconds, dx/dt = 0.2pi rad/s

differentiate both sides implicitly...

dl/dt = 180*sec^2x*dx/dt

You can get x easily by putting l = 300m and solve for your answer.

 

[freaking_out]

I have temporarily fogotten which bit the segment is

its the area in the circle bounded by the arc PQ and the chord PQ. the formula for this area is:

1/2r^2(x-sinx), x is the angle in radians

don't worry i always get mixed up with these terms

 

[McLake]

1. I'm not sure if this is right but:

Area = A = 1/2r^2(x - sinx)
dA/dx = r^2/2 - r^2/2*cosx
for max let dA/dx = 0
so r^2/2 = r^2/2*cosx
cosx = 1

x = cos^-1(x)

 

[freaking_out]

Originally posted by McLake
1. I'm not sure if this is right but:

Area = A = 1/2r^2(x - sinx)
dA/dx = r^2/2 - r^2/2*cosx
for max let dA/dx = 0
so r^2/2 = r^2/2*cosx
cosx = 1

x = cos^-1(x)

the answer at the back is x=pi

 

[kini mini]

Originally posted by freaking_out


its the area in the circle bounded by the arc PQ and the chord PQ. the formula for this area is:

1/2r^2(x-sinx), x is the angle in radians

don't worry i always get mixed up with these terms

I always got them mixed up, good thing I didn't have a go because I was leaning towards the major segment (if that's the right term!)

I get McLake's answer (x=0) by differentiating w.r.t. time . Wonder where we've stuffed up...

 

[McLake]

*Scratches head*

Hmm .... Could it be a typo?

BTW: The last line of my solution should read

x = cos^-1(1)
= 0 or 2(pi)

but you knew that ...

I'm thinking divide by two somewher, but I don't know where ...

 

[freaking_out]

Re: Re: trig. and rates of change

Originally posted by kini mini


I have temporarily fogotten which bit the segment is



An eerily familiar question! First, draw a diagram.

Let M be any point on the shoreline distinct from P.

Let angle MPL = x
Let length PM = l

Then tan x = l/180

l = 180*tan x

since the light rotates once evry 10 seconds, dx/dt = 0.2pi rad/s

differentiate both sides implicitly...

dl/dt = 180*sec^2x*dx/dt

You can get x easily by putting l = 300m and solve for your answer.

well i put l=300m but i can only get decimal answers, is there another way of doing this question???

 

[wogboy]

For question 1,

dx/dt = 0.1 rad/s

and for the area of the segment A,

A = 18x - 18sinx cm^2

therefore,

dA/dx = 18 - 18cosx cm

and since dA/dt = dA/dx * dx/dt,

dA/dt = 1.8 - 1.8cosx cm^2/s

now it asks for what value of x, is dA/dt is maximum,

this occurs when:

1.8 - 1.8cosx is maximum**

=> cosx is minumum

=> cosx = -1

=> x = pi (since x is in the range, 0<=x<=pi)

so x = pi.

** Of course you could find the first and second derivatives of 1.8 - 1.8cosx in order to find for what x is dA/dt maximum, but that's not necessary since you know that the minumum value of cosx is -1, and the minumum value of cosx leads to the maximum value of 1.8 - 1.8cosx, which is dA/dt.

 

[kini mini]

Re: Re: Re: trig. and rates of change

Originally posted by freaking_out


well i put l=300m but i can only get decimal answers, is there another way of doing this question???

I get an exact answer doing it that way.

wogboy thanks for clearing up my and McLake's error, we both seem to have forgotten that cos x could be -1.

 

[wogboy]

well i put l=300m but i can only get decimal answers, is there another way of doing this question???

If you draw up the triangle, you will see that tanx = 300/180

You should also note in particular that sec^2x = tan^2x + 1.

So sec^2x = (300/180)^2 + 1

also as pointed out before, dx/dt = 0.2pi rad/s

and since:

dl/dt = 180*sec^2x*dx/dt m/s,

dl/dl = 180 * ((300/180)^2 + 1) * 0.2pi m/s

= 180 * (34/9) * 0.2 * pi m/s

= 136pi m/s

 

[McLake]

Re: Re: Re: Re: trig. and rates of change

Originally posted by kini mini
wogboy thanks for clearing up my and McLake's error, we both seem to have forgotten that cos x could be -1.

Yes, now I feel stupid ...

 

[freaking_out]

Originally posted by McLake


Yes, now I feel stupid ...

hey i am the most stupid out of here coz i didn't know how to do the question in the first place. :mad1:

anyway thanx guys for answering my question

 

[freaking_out]

oh yeah, i forgot to ask, what topic are these questions meant to be in. i found 'em in the application of trig differentiation exercise.