Related rates - some problems =)

[eekoz]

I've been practicing related rates problems, and I want to confirm a few things.

- When a pebble is dropped in water producing a circular wave, the rate that it travels outward is the rate of the changing radius, correct? In other words, dr/dt?

Also, I have two questions I can't really figure out. Here they are:

1. The ends of a water trough are equilateral triangles whose sides are 2 m long. The length of the trough is 15m. Water is being poured into the trouth at the rate of 3m^3/h. Find the rate at which the water level is rising when the depth of the water is 1m.


2. A softball diamond has the shape of a square with sides 20 m long. Suppose a player is running from the first base to the second base at 7 m/s. Find the rate at which the distance between the player and home plate is changing when the player is 5m from the first base.


Thanks in advance for any help! Anything would be greatly appericiated =)

 

[AKG]

Right, the rate the wave travels is outwards is the rate of the changing radius, dr/dt. Show some work for the other two questions.

 

[eekoz]

Right, the rate the wave travels is outwards is the rate of the changing radius, dr/dt. Show some work for the other two questions.

Okay, so for the softball question I'm pretty sure I got that - the pyth theorem.. I got a speed of 1.7 m/s - I hope it's right (a confirmation would be good)

For the second question, I found the height of the triangles would be the sqrt(3) and the base to be 2, but not sure where to go from there. I'm not sure if it's right either.. Any confirmation about the height/base + hints?

And thank you for the confirmation about the dr/dt =)

 

[AKG]

You're right about the softball question.

There's no need to calculate the height of the trough. If x is the current height of the water, then you can express V in terms of x, and can in turn express x in terms of V. You want to find dx/dt when x = 1m, but dx/dt is just (dx/dV)(dV/dt). dV/dt is given, and dx/dV can be determined when you express x in terms of V.

By the way, the trough would be $\sqrt{3}$ meters high, as you found, but that's irrelevant. You only need to find the rate of change when the water is 1m high. Since 1m < $\sqrt{3}$m, it wouldn't matter if the trough was 1.1m or 99999m high, since if you're filling water at a constant rate, then once the water reaches 1m high, it will be rising at a particular rate which clearly doesn't depend on how much higher the trough is. For an analogy, if you drop a rock, then assuming that acceleration due to gravity is constant, the speed at which it will be falling after 1 second doesn't depend on whether you drop it from 10m above the ground or 200m above the ground, i.e. it's speed after 1s doesn't depend on how much further it has left to fall. Same goes for filling up a trough.

 

[eekoz]

You're right about the softball question.

There's no need to calculate the height of the trough. If x is the current height of the water, then you can express V in terms of x, and can in turn express x in terms of V. You want to find dx/dt when x = 1m, but dx/dt is just (dx/dV)(dV/dt). dV/dt is given, and dx/dV can be determined when you express x in terms of V.

By the way, the trough would be $\sqrt{3}$ meters high, as you found, but that's irrelevant. You only need to find the rate of change when the water is 1m high. Since 1m < $\sqrt{3}$m, it wouldn't matter if the trough was 1.1m or 99999m high, since if you're filling water at a constant rate, then once the water reaches 1m high, it will be rising at a particular rate which clearly doesn't depend on how much higher the trough is. For an analogy, if you drop a rock, then assuming that acceleration due to gravity is constant, the speed at which it will be falling after 1 second doesn't depend on whether you drop it from 10m above the ground or 200m above the ground, i.e. it's speed after 1s doesn't depend on how much further it has left to fall. Same goes for filling up a trough.

I got a rate of Sqrt(3)/10 m/h, is that correct?
Thanks for the help!

 

[AKG]

Yes, that's right.