Question
What is the rate of change of the area of the triangle formed by a 17-foot ladder leaning against a wall when the top of the ladder is falling at a rate of 2 feet per second and the bottom of the ladder is 8 feet from the wall?
Answers
GPT-4o mini
To find the rate of change of the area of the triangle formed by the 17-foot ladder against the wall, we can use geometry and related rates.
Let's denote:
- \( L = 17 \) feet (the length of the ladder),
- \( x \) = the distance from the bottom of the ladder to the wall,
- \( y \) = the height of the top of the ladder from the ground.
By the Pythagorean theorem, we have:
\[
x^2 + y^2 = L^2
\]
Substituting in the value for \( L \):
\[
x^2 + y^2 = 17^2 = 289
\]
We are given that the bottom of the ladder is \( x = 8 \) feet. We can find \( y \) when \( x = 8 \):
\[
8^2 + y^2 = 289
\]
\[
64 + y^2 = 289
\]
\[
y^2 = 289 - 64 = 225
\]
\[
y = \sqrt{225} = 15 \text{ feet}
\]
The area \( A \) of the triangle formed by the wall, the ground, and the ladder can be expressed as:
\[
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times y
\]
Substituting for \( x \) and \( y \):
\[
A = \frac{1}{2} \times 8 \times 15 = 60 \text{ square feet}
\]
Now, we want to find the rate of change of the area \( \frac{dA}{dt} \). We differentiate \( A = \frac{1}{2}xy \) using the product rule, noting that both \( x \) and \( y \) can change with time:
\[
\frac{dA}{dt} = \frac{1}{2} \left( x \frac{dy}{dt} + y \frac{dx}{dt} \right)
\]
We are given that \( \frac{dy}{dt} = -2 \) feet/second (negative because \( y \) is decreasing as the top of the ladder falls) and we will find \( \frac{dx}{dt} \) using the relationship between \( x \) and \( y \).
Differentiating the Pythagorean theorem, \( x^2 + y^2 = 289 \), we get:
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
Simplifying:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0
\]
Substituting in the known values \( x = 8 \), \( y = 15 \), and \( \frac{dy}{dt} = -2 \):
\[
8 \frac{dx}{dt} + 15(-2) = 0
\]
\[
8 \frac{dx}{dt} - 30 = 0
\]
\[
8 \frac{dx}{dt} = 30
\]
\[
\frac{dx}{dt} = \frac{30}{8} = \frac{15}{4} \text{ feet per second}
\]
Now we can calculate \( \frac{dA}{dt} \):
\[
\frac{dA}{dt} = \frac{1}{2} \left( 8(-2) + 15\left( \frac{15}{4} \right) \right)
\]
Calculating each term:
\[
\frac{dA}{dt} = \frac{1}{2} \left( -16 + \frac{225}{4} \right)
\]
Convert \(-16\) to a fraction with a denominator of 4:
\[
-16 = -\frac{64}{4}
\]
Now combining the terms:
\[
\frac{dA}{dt} = \frac{1}{2} \left( -\frac{64}{4} + \frac{225}{4} \right) = \frac{1}{2} \left( \frac{161}{4} \right) = \frac{161}{8} = 20.125 \text{ square feet per second}
\]
Thus, the rate of change of the area of the triangle is:
\[
\frac{dA}{dt} = 20.125 \text{ square feet per second}
\]
Let's denote:
- \( L = 17 \) feet (the length of the ladder),
- \( x \) = the distance from the bottom of the ladder to the wall,
- \( y \) = the height of the top of the ladder from the ground.
By the Pythagorean theorem, we have:
\[
x^2 + y^2 = L^2
\]
Substituting in the value for \( L \):
\[
x^2 + y^2 = 17^2 = 289
\]
We are given that the bottom of the ladder is \( x = 8 \) feet. We can find \( y \) when \( x = 8 \):
\[
8^2 + y^2 = 289
\]
\[
64 + y^2 = 289
\]
\[
y^2 = 289 - 64 = 225
\]
\[
y = \sqrt{225} = 15 \text{ feet}
\]
The area \( A \) of the triangle formed by the wall, the ground, and the ladder can be expressed as:
\[
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times y
\]
Substituting for \( x \) and \( y \):
\[
A = \frac{1}{2} \times 8 \times 15 = 60 \text{ square feet}
\]
Now, we want to find the rate of change of the area \( \frac{dA}{dt} \). We differentiate \( A = \frac{1}{2}xy \) using the product rule, noting that both \( x \) and \( y \) can change with time:
\[
\frac{dA}{dt} = \frac{1}{2} \left( x \frac{dy}{dt} + y \frac{dx}{dt} \right)
\]
We are given that \( \frac{dy}{dt} = -2 \) feet/second (negative because \( y \) is decreasing as the top of the ladder falls) and we will find \( \frac{dx}{dt} \) using the relationship between \( x \) and \( y \).
Differentiating the Pythagorean theorem, \( x^2 + y^2 = 289 \), we get:
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
Simplifying:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0
\]
Substituting in the known values \( x = 8 \), \( y = 15 \), and \( \frac{dy}{dt} = -2 \):
\[
8 \frac{dx}{dt} + 15(-2) = 0
\]
\[
8 \frac{dx}{dt} - 30 = 0
\]
\[
8 \frac{dx}{dt} = 30
\]
\[
\frac{dx}{dt} = \frac{30}{8} = \frac{15}{4} \text{ feet per second}
\]
Now we can calculate \( \frac{dA}{dt} \):
\[
\frac{dA}{dt} = \frac{1}{2} \left( 8(-2) + 15\left( \frac{15}{4} \right) \right)
\]
Calculating each term:
\[
\frac{dA}{dt} = \frac{1}{2} \left( -16 + \frac{225}{4} \right)
\]
Convert \(-16\) to a fraction with a denominator of 4:
\[
-16 = -\frac{64}{4}
\]
Now combining the terms:
\[
\frac{dA}{dt} = \frac{1}{2} \left( -\frac{64}{4} + \frac{225}{4} \right) = \frac{1}{2} \left( \frac{161}{4} \right) = \frac{161}{8} = 20.125 \text{ square feet per second}
\]
Thus, the rate of change of the area of the triangle is:
\[
\frac{dA}{dt} = 20.125 \text{ square feet per second}
\]