related rates:
a ladder, 12 feet long, is leaning against a wall. if the foot of the ladder slides away from the wall along level ground,what is the rate of change of the top of the ladder with respect to the distance of the foot of the ladder from the wall when the foot is 6 ft from the wall?
I don't understand what this question is asking for.
dave
Draw the 12 foot ladder up against the wall with the foot at 6 feet from the foot of the wall. Then the top is sqrt (144 - 36) feet up on the wall.
Now it starts slipping out at rate dx/dt at the bottom of the wall.
The question is, what is dy/dt, the speed of the top of the ladder down the wall.
I drew the ladder foot to the right of the wall so that dx/dt is positive :)
Of course dy/dt will change as the ladder goes down the wall even if dx/dt is constant, but the problem only asks for the vertical speed at the moment that the ladder is sqrt (108) high on the wall.
The problem is poorly worded, and I can see why you are confused. I think they want the rate at which the elevation of the top of the ladder changes (dy/dt) in terms of how fast the bottom of the ladder slides away from the wall (dx/dt), when x = 6 ft. Let x be the distance of the bottom of the ladder from the wall, and y be the elevation of the top of the ladder above the floor.
Start with
x^2 + y^2 = 12^2 = 144
Both x and y are functions of t. Differentiate both sides of the abovew equation with respect to t.
2x dx/dt = -2y dy/dt
dy/dt = - (x/y) dx/dt
When x = 6, y = sqrt (144-36)= 10.39, so
dy/dt = -0.577 dx/dt
From the foot of the building I look up at an angle 22 from horizontal in order to see the top tree.from the top building 150 m abovethe ground level i have look down At 50 from the hirizontal in order to see the top.
A.what is the heightof the tree?
B.how far from the building is the tree
I can't understand
This question is asking for the rate of change of the top of the ladder with respect to the distance of the foot of the ladder from the wall. In other words, it wants to know how fast the height of the ladder is changing as the foot of the ladder moves away from the wall.
To solve this problem, we can use the concept of related rates. Related rates involve finding the rate at which one quantity is changing with respect to another related quantity.
In this case, we can set up a right triangle with the wall, the ground, and the ladder as its sides. The ladder is the hypotenuse, and the distance of the foot of the ladder from the wall is one of the sides.
Let's call the distance of the foot of the ladder from the wall "x" and the height of the ladder "y". We are given that the length of the ladder is 12 feet, so we can write the equation of the right triangle as:
x^2 + y^2 = 12^2
To find the rate of change of the height of the ladder with respect to the distance from the wall, we need to differentiate both sides of the equation with respect to time. Let's assume that both x and y are functions of time, so we have:
(d/dt) [x^2 + y^2] = (d/dt) [12^2]
Now, let's differentiate both sides of the equation using the chain rule. The chain rule states that if y = f(g(x)), then the derivative of y with respect to x is given by:
(dy/dx) = (dy/dg) * (dg/dx)
Using the chain rule, we have:
2x(dx/dt) + 2y(dy/dt) = 0
Now, we need to find the values of x, y, and dx/dt when the foot of the ladder is 6 feet from the wall. We are given that x = 6, and we need to find y and dx/dt.
To find y, we can use the equation of the right triangle:
6^2 + y^2 = 12^2
Simplifying this equation, we get:
y^2 = 12^2 - 6^2
y^2 = 144 - 36
y^2 = 108
y = √108
Now, let's differentiate both sides of the equation x^2 + y^2 = 12^2 with respect to time:
2x(dx/dt) + 2y(dy/dt) = 0
Plugging in the values we found earlier, we have:
2(6)(dx/dt) + 2(√108)(dy/dt) = 0
Simplifying this equation, we get:
12(dx/dt) + 2(√108)(dy/dt) = 0
Now, we can solve this equation for dy/dt, which represents the rate of change of the height of the ladder with respect to the distance from the wall.