Related Rates: right triangle w/ fixed base, incr. height; find change rate of angle

[nanluo]

15. Consider the right triangle where the length x of the vertical leg (that is, the height) is increasing at the rate of 50 units per second, the horizontal base has a fixed length of 10 units, and the base angle "theta" is opposite the vertical leg.

In radians per second, what is the rate of change of the angle "theta" when x = 10?

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Hi I'm looking for help for a related rates word problem. I believe it's similar to the falling ladder question but I just can't seem to get it. Thanks!

 

[tkhunny]

Well, you must start with \(\displaystyle x = f(\theta)\;or\;\theta = g(x)\). Do that.

 

[Steven G]

15. Consider the right triangle where the length x of the vertical leg (that is, the height) is increasing at the rate of 50 units per second, the horizontal base has a fixed length of 10 units, and the base angle "theta" is opposite the vertical leg.

In radians per second, what is the rate of change of the angle "theta" when x = 10?

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Hi I'm looking for help for a related rates word problem. I believe it's similar to the falling ladder question but I just can't seem to get it. Thanks!

These problems start to look all the same once you write down what is given and what is being asked for.

I'll start you off with the following: The distance x is increasing at the rate of 50 units per second translates to dx/dt = 50 units/sec
What is the rate of change of theta when x = 10 translates to d theta/dt = ??? when x=10

So what variables are we working with? Answer theta and x.
Any formula connecting theta and x?? You find that one. Just look at the diagram.

Now that you have a formula using theta and x take the derivative of it wrt t and solve that equation for d theta/dt.....