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

[chengeto]

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)

 

[Deleted member 4993]

Re: Trig related rates

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

\(\displaystyle S \, = \, r\cdot \theta \, = \, 2\cdot r\cdot \alpha\)

and

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

and

\(\displaystyle \frac{dD}{dt} \, = \, v \, = \, constant\)

you need to find

\(\displaystyle \frac{dS}{dt} \, = \, ???\)

Now continue.....

 

[chengeto]

Re: Trig related rates

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

\(\displaystyle S \, = \, r\cdot \theta \, = \, 2\cdot r\cdot \alpha\)

and

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

and

\(\displaystyle \frac{dD}{dt} \, = \, v \, = \, constant\)

you need to find

\(\displaystyle \frac{dS}{dt} \, = \, ???\)

Now continue.....

\(\displaystyle S=20\alpha\)

\(\displaystyle \alpha=arctan\frac{D}{10}\)

\(\displaystyle \frac{d\alpha}{dt}=\frac{1}{1+\frac{D}{10}}\frac{1}{500}\)

\(\displaystyle \frac{dS}{dt}= 20\frac{d\alpha}{dt}\)

Is this correct ?

 

[Deleted member 4993]

Re: Trig related rates

\(\displaystyle S=20\alpha\)

\(\displaystyle \alpha=arctan\frac{D}{10}\)

\(\displaystyle \frac{d\alpha}{dt}=\frac{1}{1+\frac{D}{10}}\frac{1}{500}\) ...........Incorrect

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

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\(\displaystyle \frac{dS}{dt}= 20\frac{d\alpha}{dt}\)

Is this correct ?

 

[galactus]

Here is my solution. I worked it out and then forgot about it.

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

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

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

\(\displaystyle \frac{ds}{dt}=20\frac{d{\alpha}}{dt}\)..............[1]

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

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

\(\displaystyle sec^{2}({\alpha})\frac{d{\alpha}}{dt}=\frac{1}{10}\frac{dD}{dt}\)

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

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

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

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