Help with trigonometry.

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KFS

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In Example 1: I don't understand the procedure of this exercise. I do know why the derivative of the angle with respect to time is 2π/30. What I don't understand is the rest. It says x=10tanθ, how is this possible? It gives me, and the computer, by tanθ=opposite/adjacent, x=10/tanθ, not 10tanθ. What I don't understand either is why both derivatives (dx/dθ) and (dθ/dt) are multiplying. Why is x=8 choosen? Why not any other number?
Thanks for the help.
 

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I posted a general solution to this kind of problem here:

Related Rates Part IV

This whole section is kicking my butt. The beacon on a lighthouse makes a revolution every 20 secs. The beacon is 300ft away from nearest point P on the shoreline. Find the rate at which the ray of light moves along the shoreline at a point 200ft from point P. Let the lighthouse be the apex of...
www.freemathhelp.com

It should answer your questions.
 
MarkFL said:
I posted a general solution to this kind of problem here:

Related Rates Part IV

This whole section is kicking my butt. The beacon on a lighthouse makes a revolution every 20 secs. The beacon is 300ft away from nearest point P on the shoreline. Find the rate at which the ray of light moves along the shoreline at a point 200ft from point P. Let the lighthouse be the apex of...
www.freemathhelp.com

It should answer your questions.
Click to expand...
Thank you, I appreciate this.
 
I wonder if you are misreading the problem. Note where [MATH]\theta[/MATH] is, and where x is.
 
KFS said:
In Example 1: I don't understand the procedure of this exercise. I do know why the derivative of the angle with respect to time is 2π/30. What I don't understand is the rest. It says x=10tanθ, how is this possible? It gives me, and the computer, by tanθ=opposite/adjacent, x=10/tanθ, not 10tanθ. What I don't understand either is why both derivatives (dx/dθ) and (dθ/dt) are multiplying. Why is x=8 choosen? Why not any other number?
Thanks for the help.
Click to expand...
Why is x=8 choosen? Why not any other number? That is not really possible to answer. The author simple pick P to be 8km from Q as shown in the diagram. In other words it was given. Possibly Dr P is correct that you not realizing that x is the horizontal distance from point Q?

I don't understand either is why both derivatives (dx/dθ) and (dθ/dt) are multiplying. As a calculus 1 student you can think of the dθ's cancelling out so (dx/dθ)* (dθ/dt)= (dx/dt) and you need (dx/dt) as that is what they are asking for!

Yes, tanθ=opposite/adjacent but where the opposite and adjacent are depends on where the angle θ is which is in the top right corner. Re-read the problem to see why θ is located there. tanθ = x/10 so x = 10tanθ.

Please let us know if you are still stuck
 
Jomo said:
Why is x=8 choosen? Why not any other number? That is not really possible to answer. The author simple pick P to be 8km from Q as shown in the diagram. In other words it was given. Possibly Dr P is correct that you not realizing that x is the horizontal distance from point Q?

I don't understand either is why both derivatives (dx/dθ) and (dθ/dt) are multiplying. As a calculus 1 student you can think of the dθ's cancelling out so (dx/dθ)* (dθ/dt)= (dx/dt) and you need (dx/dt) as that is what they are asking for!

Yes, tanθ=opposite/adjacent but where the opposite and adjacent are depends on where the angle θ is which is in the top right corner. Re-read the problem to see why θ is located there. tanθ = x/10 so x = 10tanθ.

Please let us know if you are still stuck
Click to expand...
I'm not stuck anymore. Thanks for your help. I thought dy/dx was just a notation that you "couldn't manipulate it". I've done many things involving differential calculus and a few things involving integral calculus (Riemann sums, Telescoping Sums, Fundamental Theorem of Calculus...), but word problems are always the hardest for me. I just get nervous and think I'll fail, like in this case. Anyway, thanks for the help.
 
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