Question
A 5-foot board slides down a wall. At the instant the bottom end is 4 feet from the wall, the other end is moving down the wall at a rate of 2 feet per second. How fast is the bottom end sliding along the ground at that moment?
Answers
GPT-4o mini
To solve the problem, we can use related rates in calculus. Let's define the variables involved:
- Let \( y \) be the height of the top of the board from the ground at the wall.
- Let \( x \) be the distance from the bottom of the board to the wall.
- The length of the board is 5 feet.
By the Pythagorean theorem, we have:
\[
x^2 + y^2 = 5^2 = 25.
\]
We're given the following information:
- At a specific moment, \( x = 4 \) feet.
- The rate of change of \( y \) (the top end moving down) is given as \( \frac{dy}{dt} = -2 \) feet per second (the negative sign is used because \( y \) is decreasing).
We want to find \( \frac{dx}{dt} \), the rate at which the bottom end of the board slides along the ground.
First, let's differentiate the equation \( x^2 + y^2 = 25 \) with respect to time \( t \):
\[
\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25).
\]
Using the chain rule:
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0.
\]
We can simplify by dividing everything by 2:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0.
\]
Now we can solve for \( \frac{dx}{dt} \):
\[
x \frac{dx}{dt} = -y \frac{dy}{dt},
\]
\[
\frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt}.
\]
Next, we need to find the value of \( y \) when \( x = 4 \) feet. We can substitute \( x = 4 \) into the Pythagorean theorem:
\[
4^2 + y^2 = 25,
\]
\[
16 + y^2 = 25,
\]
\[
y^2 = 9,
\]
\[
y = 3 \text{ feet.}
\]
Now substitute \( x \), \( y \), and \( \frac{dy}{dt} \) into the related rates equation:
\[
\frac{dx}{dt} = -\frac{3}{4} \cdot (-2),
\]
\[
\frac{dx}{dt} = \frac{3 \cdot 2}{4} = \frac{6}{4} = \frac{3}{2}.
\]
Thus, the bottom end of the board slides along the ground at a rate of:
\[
\boxed{1.5} \text{ feet per second.}
\]
- Let \( y \) be the height of the top of the board from the ground at the wall.
- Let \( x \) be the distance from the bottom of the board to the wall.
- The length of the board is 5 feet.
By the Pythagorean theorem, we have:
\[
x^2 + y^2 = 5^2 = 25.
\]
We're given the following information:
- At a specific moment, \( x = 4 \) feet.
- The rate of change of \( y \) (the top end moving down) is given as \( \frac{dy}{dt} = -2 \) feet per second (the negative sign is used because \( y \) is decreasing).
We want to find \( \frac{dx}{dt} \), the rate at which the bottom end of the board slides along the ground.
First, let's differentiate the equation \( x^2 + y^2 = 25 \) with respect to time \( t \):
\[
\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25).
\]
Using the chain rule:
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0.
\]
We can simplify by dividing everything by 2:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0.
\]
Now we can solve for \( \frac{dx}{dt} \):
\[
x \frac{dx}{dt} = -y \frac{dy}{dt},
\]
\[
\frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt}.
\]
Next, we need to find the value of \( y \) when \( x = 4 \) feet. We can substitute \( x = 4 \) into the Pythagorean theorem:
\[
4^2 + y^2 = 25,
\]
\[
16 + y^2 = 25,
\]
\[
y^2 = 9,
\]
\[
y = 3 \text{ feet.}
\]
Now substitute \( x \), \( y \), and \( \frac{dy}{dt} \) into the related rates equation:
\[
\frac{dx}{dt} = -\frac{3}{4} \cdot (-2),
\]
\[
\frac{dx}{dt} = \frac{3 \cdot 2}{4} = \frac{6}{4} = \frac{3}{2}.
\]
Thus, the bottom end of the board slides along the ground at a rate of:
\[
\boxed{1.5} \text{ feet per second.}
\]