6. A 10-meter ladder is leaning against the wall of a building. It begins to slide down the wall. When the end of the ladder on the ground is 6 meters from the building, that end is moving at 1 meter per second. At that time how fast is the end against the wall moving down?
6. A 10-meter ladder is leaning against the wall of a building. It begins to slide down the wall. When the end of the ladder on the ground is 6 meters from the building, that end is moving at 1 meter per second. At that time how fast is the end against the wall moving down?
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![**Problem 6: Related Rates with a Ladder**
A 10-meter ladder is leaning against the wall of a building. It begins to slide down the wall. When the end of the ladder on the ground is 6 meters from the building, that end is moving at 1 meter per second. At that time, how fast is the end against the wall moving down?
**Explanation:**
This problem involves related rates. The ladder, wall, and ground form a right triangle where:
- The ladder is the hypotenuse, with a constant length of 10 meters.
- The distance from the wall to the bottom of the ladder is \( x \) meters.
- The height from the ground to the top of the ladder on the wall is \( y \) meters.
Given:
- \(\frac{dx}{dt} = 1 \, \text{m/s}\) (the rate at which the bottom of the ladder moves away from the wall).
- At the instance when \( x = 6 \, \text{m}\).
We need to find \(\frac{dy}{dt}\), the rate at which the top of the ladder is descending.
**Using the Pythagorean theorem:**
\[ x^2 + y^2 = 10^2 \]
Differentiating both sides with respect to \( t \):
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Substitute known values:
\[ 2(6)(1) + 2y \frac{dy}{dt} = 0 \]
\[ 12 + 2y \frac{dy}{dt} = 0 \]
\[ 2y \frac{dy}{dt} = -12 \]
To find \( y \), use:
\[ x^2 + y^2 = 100 \]
\[ 6^2 + y^2 = 100 \]
\[ 36 + y^2 = 100 \]
\[ y^2 = 64 \]
\[ y = 8 \]
Substitute \( y = 8 \) back:
\[ 2(8) \frac{dy}{dt} = -12 \]
\[ 16 \frac{dy}{dt} = -12 \]
\[ \frac{dy}{dt} = -\frac{12}{16} \]
\[ \frac{dy}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7567ae90-f1e1-4ad2-8c63-c8b884d687f9%2Fabbf64e8-a9c4-427f-8cb8-dbcfc8e3f36c%2Fska8mdf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 6: Related Rates with a Ladder**
A 10-meter ladder is leaning against the wall of a building. It begins to slide down the wall. When the end of the ladder on the ground is 6 meters from the building, that end is moving at 1 meter per second. At that time, how fast is the end against the wall moving down?
**Explanation:**
This problem involves related rates. The ladder, wall, and ground form a right triangle where:
- The ladder is the hypotenuse, with a constant length of 10 meters.
- The distance from the wall to the bottom of the ladder is \( x \) meters.
- The height from the ground to the top of the ladder on the wall is \( y \) meters.
Given:
- \(\frac{dx}{dt} = 1 \, \text{m/s}\) (the rate at which the bottom of the ladder moves away from the wall).
- At the instance when \( x = 6 \, \text{m}\).
We need to find \(\frac{dy}{dt}\), the rate at which the top of the ladder is descending.
**Using the Pythagorean theorem:**
\[ x^2 + y^2 = 10^2 \]
Differentiating both sides with respect to \( t \):
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Substitute known values:
\[ 2(6)(1) + 2y \frac{dy}{dt} = 0 \]
\[ 12 + 2y \frac{dy}{dt} = 0 \]
\[ 2y \frac{dy}{dt} = -12 \]
To find \( y \), use:
\[ x^2 + y^2 = 100 \]
\[ 6^2 + y^2 = 100 \]
\[ 36 + y^2 = 100 \]
\[ y^2 = 64 \]
\[ y = 8 \]
Substitute \( y = 8 \) back:
\[ 2(8) \frac{dy}{dt} = -12 \]
\[ 16 \frac{dy}{dt} = -12 \]
\[ \frac{dy}{dt} = -\frac{12}{16} \]
\[ \frac{dy}{
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