**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}{

Given the length of the ladder is, 10 - meter. The rate at which the end of the ladder moving is 1 meter per second The distance between the end and the building is 6 meters when the rate is 1 meter per second. The given data's can be shown below, Let the length base side ( ladder and building) be x and the altitude( distance from the bottom of building to the top of ladder be y.To find the rate at which the end against the wall moving down: From this data's dxdt=1 m/sec when x=6m. To find dydt, By Pythagoras theorem x2+y2=102 Differentiate both sides with respect to to t, 2xdxdt+2ydydt=0 ⇒dydt=-xydxdt When x=6 meter, y=102-62=4 meters Thus, dydt=-64×1=-64=-32m/sec. Thus, the top of the ladder is sliding down at the rate of 32m/sec.