How can I calculate the rate of change between a rising balloon and a cyclist?

[jacy]

Hello,
This is what am trying to solve

A balloon is rising at the rate of 5ft/s. A boy is cycling along a straight road at a speed of 15ft/s.when he passes under the balloon it is 45 ft above him.
How fast is the distance between the balloon and the boy increasing 3 seconds later

Please let me know if am on the right path
s^2=x^2+y^2

dy/dt = 5
dx/dt = 15
y= 45

so

2s ds/dt = 2x dx/dt + 2y dy/dt

is this correct but also we can conclude that ds/dt = 15.81,
but i think we have to find ds/dt.
In the above equation am getting two unkowns that's were am getting stuck. Please help, thanks.

 

[0rthodontist]

The "y = 45" you should write as y(0) = 45 since y is really a function. In the equation you have last, 2s ds/dt = 2x dx/dt + 2y dy/dt, you have all the information you need. You can find s, x, and y by simple algebra and geometry, and you already know dx/dt and dy/dt. Then you can solve easily.

 

[jacy]

The "y = 45" you should write as y(0) = 45 since y is really a function. In the equation you have last, 2s ds/dt = 2x dx/dt + 2y dy/dt, you have all the information you need. You can find s, x, and y by simple algebra and geometry, and you already know dx/dt and dy/dt. Then you can solve easily.

Thanks for replying. In this problem since its given to us that the cyclist is traveling at 15 ft/s, so in 3 seconds he will cover 45 ft and this will be the value of x. Now i have x, and y so i can find s and that will be s = 63.64 ft.

Now i can plug everything in the equation

ds/dt = 1/s (x dx/dt + y dy/dt)

so am getting ds/dt = 14.14 ft/s

please let me know if this is correct, sorry bout 15.81.

 

[0rthodontist]

Not quite, you don't have the right value for s or y. Remember the balloon is moving too.

sorry bout 15.81.

My attitude is, it's far better to make errors at first but be right in the end than say nothing.

 

[jacy]

Not quite, you don't have the right value for s or y. Remember the balloon is moving too.


My attitude is, it's far better to make errors at first but be right in the end than say nothing.

I used pythagorean theorem to get 15.81. I was thinking 15 to be the base and 5 to be the adjacent side.

Y= 45 ft high only when the cyclist is right under it. As the cyclist is moving away the balloon keeps on rising so the y value is changing. Balloon is rising at the rate of 5 ft/s, so this is dy/dt and 15 ft/s is dx/dt. My value of x is correct right. The cyclist is moving at 15 ft/s and in 3 seconds he would cover 45 ft. When the cyclist is 45 ft away from the balloon the height of the balloon will be 4 Y = 45 + y true. Thanks for looking into this problem.

 

[0rthodontist]

The height of the balloon will be 60 feet. It is at 45 feet at time 0 and it moves with a speed of 5 ft/s over a period of 3 seconds.

 

[jacy]

The height of the balloon will be 60 feet. It is at 45 feet at time 0 and it moves with a speed of 5 ft/s over a period of 3 seconds.

Thanks orthodontist for ur help, am getting the answer