Trig related rates: hemispherical swimming pool w/ r = 10m

chengeto

New member
Joined
Feb 28, 2009
Messages
49
A hemi-spherical swimming pool has a radius of 10 metres and is completely full of water. A stone is dropped into the pool, at the centre of the surface of the water, and falls vertically at a constant rate of one metre every five seconds. A light at the edge of the pool casts a shadow of the stone on the opposite side of the pool, as shown in the accompanying diagram. Find the rate at which the shadow is moving along the side of the pool at the instant when the stone lies at a depth of 5 metres. [HINT: You may use the fact that, in the accompanying diagram,? = 2? .]

Attempt to solution: I am stuck on how do l make up the relation or equation for this problem. Any tips and advise will be greatly appreciated.

( Guys l am sorry for posting so much on one day l got an exam tomorrow so l am just doing questions from previous exam papers)
 

Attachments

  • related rates.jpg
    related rates.jpg
    11.3 KB · Views: 353
Re: Trig related rates

chengeto said:
A hemi-spherical swimming pool has a radius of 10 metres and is completely full of water. A stone is dropped into the pool, at the centre of the surface of the water, and falls vertically at a constant rate of one metre every five seconds. A light at the edge of the pool casts a shadow of the stone on the opposite side of the pool, as shown in the accompanying diagram. Find the rate at which the shadow is moving along the side of the pool at the instant when the stone lies at a depth of 5 metres. [HINT: You may use the fact that, in the accompanying diagram,? = 2? .]


Attempt to solution:

I am stuck on how do l make up the relation or equation for this problem. Any tips and advise will be greatly appreciated.


( Guys l am sorry for posting so much on one day l got an exam tomorrow so l am just doing questions from previous exam papers)
use

S=rθ=2rα\displaystyle S \, = \, r\cdot \theta \, = \, 2\cdot r\cdot \alpha

and

tan(α)=Dr\displaystyle \tan(\alpha) \, = \, \frac{D}{r}

and

dDdt=v=constant\displaystyle \frac{dD}{dt} \, = \, v \, = \, constant

you need to find

dSdt=???\displaystyle \frac{dS}{dt} \, = \, ???

Now continue.....
 
Re: Trig related rates

Subhotosh Khan said:
chengeto said:
A hemi-spherical swimming pool has a radius of 10 metres and is completely full of water. A stone is dropped into the pool, at the centre of the surface of the water, and falls vertically at a constant rate of one metre every five seconds. A light at the edge of the pool casts a shadow of the stone on the opposite side of the pool, as shown in the accompanying diagram. Find the rate at which the shadow is moving along the side of the pool at the instant when the stone lies at a depth of 5 metres. [HINT: You may use the fact that, in the accompanying diagram,? = 2? .]


Attempt to solution:

I am stuck on how do l make up the relation or equation for this problem. Any tips and advise will be greatly appreciated.


( Guys l am sorry for posting so much on one day l got an exam tomorrow so l am just doing questions from previous exam papers)
use

S=rθ=2rα\displaystyle S \, = \, r\cdot \theta \, = \, 2\cdot r\cdot \alpha

and

tan(α)=Dr\displaystyle \tan(\alpha) \, = \, \frac{D}{r}

and

dDdt=v=constant\displaystyle \frac{dD}{dt} \, = \, v \, = \, constant

you need to find

dSdt=???\displaystyle \frac{dS}{dt} \, = \, ???

Now continue.....

S=20α\displaystyle S=20\alpha

α=arctanD10\displaystyle \alpha=arctan\frac{D}{10}

dαdt=11+D101500\displaystyle \frac{d\alpha}{dt}=\frac{1}{1+\frac{D}{10}}\frac{1}{500}

dSdt=20dαdt\displaystyle \frac{dS}{dt}= 20\frac{d\alpha}{dt}

Is this correct ?
 
Re: Trig related rates

chengeto said:
S=20α\displaystyle S=20\alpha

α=arctanD10\displaystyle \alpha=arctan\frac{D}{10}

dαdt=11+D101500\displaystyle \frac{d\alpha}{dt}=\frac{1}{1+\frac{D}{10}}\frac{1}{500} ...........Incorrect

________________________________________

dtan1(ax)dx=a1+(ax)2\displaystyle \frac{d\tan^{-1}(ax)}{dx} \, = \frac{a}{1+(ax)^2}

____________________________________________


dSdt=20dαdt\displaystyle \frac{dS}{dt}= 20\frac{d\alpha}{dt}

Is this correct ?
 
Here is my solution. I worked it out and then forgot about it.

Remember the old circle geometry theorem which says that θ=2α\displaystyle {\theta}=2{\alpha}, as in the hint?.

We know that s=rθ\displaystyle s=r{\theta}. Therefore, s=2rα\displaystyle s=2r{\alpha}

We are told that r=10. So, it is a constant.

dsdt=20dαdt\displaystyle \frac{ds}{dt}=20\frac{d{\alpha}}{dt}..............[1]

From the problem statement, ds/dt is what we want. So, we have to find dαdt\displaystyle \frac{d{\alpha}}{dt}, plug it in and we have it.

tan(α)=D10\displaystyle tan({\alpha})=\frac{D}{10}

sec2(α)dαdt=110dDdt\displaystyle sec^{2}({\alpha})\frac{d{\alpha}}{dt}=\frac{1}{10}\frac{dD}{dt}

we are told that dD/dt=1/5 meter per second.

When D=5, then sec(tan1(12))dαdt=11015\displaystyle sec(tan^{-1}(\frac{1}{2}))\frac{d{\alpha}}{dt}=\frac{1}{10}\cdot\frac{1}{5}

Solving, we find that dαdt=2125\displaystyle \frac{d{\alpha}}{dt}=\frac{2}{125}

Plug into the [1] and we get dsdt=825=.32   msec\displaystyle \frac{ds}{dt}=\frac{8}{25}=.32 \;\ \frac{m}{sec}
 
Top