Help solving inverse and joint variation word problems

[theels315]

I have to solve 2 word problems one is a joint variation problem and the other is an inverse variation problem..The Joint variation problem is as follows:

1) The heat loss of a glass window varies jointly as the window's area and the difference between the outside and inside temperatures. A window 3 feet wide by 6 feet long loses 1200 Btu per hour when the temperature outside is 20 degrees colder than the temperature inside. Find the heat loss through a glass window that is 6 feet wide by 9 feet long when the temperature outside is 10 degrees colder that the temperature inside.

The inverse variation problem is as follows:

2) Sound intensity varies inversely as the square root of the distance from the sound source. If you are in a movie theater and you change your seat to one that is twice as far from the speakers, how does the new sound intensity compare to that of your original seat?

I know the first step in solving these problems is to write an equation, however I tried with the joint variation problem and I was told it was incorrect..The equation I put was H(heat)=ka(area)(temperature).
And for the inverse I put the square root of two,and was told it was incorrect...Please help me I need to have these in the morning and I've been working on them for a few days...Thanks so much in advance.

 

[galactus]

The heat loss of a glass window varies jointly as the window's area and the difference between the outside and inside temperatures. A window 3 feet wide by 6 feet long loses 1200 Btu per hour when the temperature outside is 20 degrees colder than the temperature inside. Find the heat loss through a glass window that is 6 feet wide by 9 feet long when the temperature outside is 10 degrees colder that the temperature inside.

Let H=heat loss, a=area, d=difference in temp.

I would think it would be \(\displaystyle \L\\H=kad\)

From the given info: \(\displaystyle 1200=k(18)(20)\)

Now, solve for k and use it in the formula with the data for the 9 by 6 window to find the BTU's lost.

 

[theels315]

Let H=heat loss, a=area, d=difference in temp.

I would think it would be \(\displaystyle \L\\H=kad\)

From the given info: \(\displaystyle 1200=k(18)(20)\)

Now, solve for k and use it in the formula with the data for the 9 by 6 window to find the BTU's lost.

Thanks so much for all of your help! I really do appreciate it. Now how would I set up the 2nd equation, its difficult to me, being that there are no numbers. Once again I appreciate it.

 

[theels315]

For the "intensity" problem, the book says the correct answer is "1/4 of what is was originally"..What was plugged in to obtain this answer?

For the joint variation problem, I looked in the back of my math book and thanks to your help, I FINALLY see how to solve the problem. Thanks so much.

 

[theels315]

Set up your inverse proportion.

Let i=intensity of sound, d=distance.

\(\displaystyle \L\\i=\frac{k}{\sqrt{d}}\)

You are given a distance twice as far in order to find k.

\(\displaystyle \L\\i=\frac{k}{\sqrt{2d}}\)

\(\displaystyle \L\\k=i\sqrt{2d}\)

Does anyone know how to do this? I'm still trying to figure it out.

 

[galactus]

I believe that should be "varies inversely as the square of the distance".

\(\displaystyle \L\\i=\frac{k}{(d)^{2}}\)

\(\displaystyle \L\\id^{2}=k\)

\(\displaystyle \L\\i=\frac{id^{2}}{(2d)^{2}}=\frac{i}{4}\)

In other words, it is 1/4th as intense.

 

[stapel]

Does anyone know how to do this? I'm still trying to figure it out.

You've been given the complete set-up and explanation. Where are you stuck?

Please reply showing your work. Thank you.

Eliz.

 

[theels315]

Does anyone know how to do this? I'm still trying to figure it out.

You've been given the complete set-up and explanation. Where are you stuck?

Please reply showing your work. Thank you.

Eliz.

Ok, now I'm stuck as to what happened to the k in the last equation and how the 2d^2 was put into the denominator.

 

[Gene]

The second equation says what k equals. Then you substitute that in the third (where the distance is 2d) the k is gone. The