Rate of change of the area of the rebgion

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Homework Help Overview

The problem involves a board sliding down a wall, creating a triangular area between the board, the ground, and the wall. The specific question is about the rate of change of this area at a given moment when the board's position is defined.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the geometric relationship of the board's position and the area it creates, referencing the triangle formed. There are attempts to apply the Pythagorean theorem and derivatives to find the rate of change of the area.

Discussion Status

Some participants are questioning the assumptions about the shape formed by the board and the ground, while others are clarifying the relationships between the variables involved. There is an acknowledgment of potential errors in the application of derivatives and variable consistency.

Contextual Notes

There is a focus on understanding how the dimensions of the triangle change as the board slides, indicating that the variables should not be treated as fixed values but rather as changing quantities throughout the motion.

fifaking7
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Homework Statement



A board 5 feet long slides down a wall. at the instant the bottom end is 4 feet from the wall, the other end is moving down the wall the rate of 2 feet per second.at that moment how fast is the area of the region between the board, the ground and the wall changing?

Homework Equations


a=1/2bh


The Attempt at a Solution


x^2 + 4^2= 5^2
x=3
da/dt= 1/2b(db/dt)(dh/dt)

I think i started it up wrong and I don't know where to go next..
 
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fifaking7 said:

Homework Statement



A board 5 feet long slides down a wall. at the instant the bottom end is 4 feet from the wall, the other end is moving down the wall the rate of 2 feet per second.at that moment how fast is the area of the region between the board, the ground and the wall changing?

Homework Equations


a=1/2bh


The Attempt at a Solution


x^2 + 4^2= 5^2
x=3
da/dt= 1/2b(db/dt)(dh/dt)

I think i started it up wrong and I don't know where to go next..
What is the shape of the region below the board and above the ground at the moment under consideration?
 
just a normal triangle drawn on a line like a typical triangle leaning against a wall problem.
 
Neither end of the board is touching the ground, so it's not a triangle.
 
fifaking7 said:

Homework Equations


a=1/2bh

da/dt= 1/2b(db/dt)(dh/dt)

Have another look at that derivative. It looks like the product rule hasn't been applied properly. Also one of those b/h is x. You'll have to stay consistent with your variable names.
 


http://img231.imageshack.us/img231/5130/scan0060kd.jpg

that is how it looks
 
Last edited by a moderator:


fifaking7 said:
http://img231.imageshack.us/img231/5130/scan0060kd.jpg

that is how it looks
You're right. I misread the problem.

When you said x=3, that should be h=3 .

In general, how are b and h related, considering that they're legs of a right triangle?
 
Last edited by a moderator:
i said b= 4ft and h=3 ft
 
fifaking7 said:
i said b= 4ft and h=3 ft
That's what b & h are at the instant that h = 4 ft, but how are they related in general?

(Use the Pythagorean theorem.)
 
  • #10
fifaking7 said:
i said b= 4ft and h=3 ft

The crucial point that you seem to be missing is that you need to figure out what is happening as the board slides down the wall, so h moves from more than 3 ft to less than 3 ft (and, at the same time, b moves from less than 4 ft to more than 4 ft). When that is happening, the area of the triangle is changing, and that is what you are supposed to be reckoning. So, you need to let b and h be variables, not fixed numbers.

RGV
 
  • #11
SammyS said:
That's what b & h are at the instant that h = 4 ft, but how are they related in general?

(Use the Pythagorean theorem.)
Of course, that's a typo !

It should have said:
That's what b & h are at the instant that b = 4 ft, but how are they related in general?
 

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