Relating Volume to Height||Trough Equation

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lroscios8232

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Hi everyone, I'm new here. This is a related rates problem that I completely flunked on my last test.

6. A 400 cm long watering trough has ends that are equilateral triangles whose sides are 60 cm. To the nearest hundredth, find the rate at which water is draining out of the bottom of the trough if the depth of the water is dropping at 3 cm/s when the depth is 8 cm.

Now, I understand the basis of related rates is to identify what you're given, what you're finding, and then connecting the two via an equation. So from what I understand dh/dt is -3cm/s when h=8, and we're looking for dV/dt, but I'm not sure how to proceed from there.

Any help would be appreciated.

Thanks,
Reizo
 
lroscios8232 said:
Hi everyone, I'm new here. This is a related rates problem that I completely flunked on my last test.

6. A 400 cm long watering trough has ends that are equilateral triangles whose sides are 60 cm. To the nearest hundredth, find the rate at which water is draining out of the bottom of the trough if the depth of the water is dropping at 3 cm/s when the depth is 8 cm.

Now, I understand the basis of related rates is to identify what you're given, what you're finding, and then connecting the two via an equation. So from what I understand dh/dt is -3cm/s when h=8, and we're looking for dV/dt, but I'm not sure how to proceed from there.

Any help would be appreciated.

Thanks,
Reizo
Click to expand...
Did you draw a sketch of the trough?

If yes - please share the sketch with proper designations.
 
Define "depth".

Find Volume = f(depth) -- Use your best geometry.

You're almost done.
 
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