A ladder 29 feet long leans against a wall and the foot of the ladder is sliding away at a constant rate of 3 feet/sec. Meanwhile, a firefighter is climbing up the ladder at a rate of 2 feet/sec. When the firefighter has climbed up 6 feet of the ladder, the ladder makes an angle of π/3 with the ground. Answer the two related rates questions below. (Hint: Use two carefully labeled similar right triangles.)
I think that the θ is just a way of giving us the slope of the ladder. If the firefighter has climbed a distance z to a height h, then we have
x^2+y^2 = 29^2
w^2+(y-h)^2 = (29-z)^2
when θ=π/3, x = 29/2 and y=29√3/2
w = (29-z)/2 and h = z√3/2
when z=6, then,
w = 23/2 and h=3√3
2x dx/dt + 2y dy/dt = 0
2(29/2)(3) + 2(29√3/2) dy/dt = 0
dy/dt = -√3
See if you can finish it off, using the other equation.
To solve the given problem, we'll start by creating a diagram and labeling the relevant quantities. Let's call the distance between the foot of the ladder and the wall "x" and the distance from the top of the ladder to the ground "y."
We are given that the ladder is 29 feet long, so we can write the Pythagorean theorem to relate x and y:
x^2 + y^2 = 29^2
Now, let's differentiate both sides of this equation with respect to time to find the relationship between the rates of change:
2x(dx/dt) + 2y(dy/dt) = 0
We know that dx/dt is given as -3 ft/sec since the foot of the ladder is sliding away at a constant rate of 3 feet/sec. Let's plug in this value:
2x(-3) + 2y(dy/dt) = 0
Since we're interested in finding dy/dt, let's solve for it:
-6x + 2y(dy/dt) = 0
dy/dt = (6x) / (2y)
dy/dt = 3x / y
Now, we need to find the values of x and y when the firefighter has climbed up 6 feet of the ladder and the ladder makes an angle of π/3 with the ground.
Let's consider the right triangle formed by the ladder, the wall, and the ground. Since the ladder makes an angle of π/3 with the ground, the opposite side has a length of y = 6 feet, which is the distance the firefighter has climbed.
To find x, we can use trigonometry. We know that tan(π/3) = y / x:
tan(π/3) = 6 / x
√3 = 6 / x
3x = 6
x = 2 feet
Plugging the values of x = 2 feet and y = 6 feet into our equation for dy/dt, we can find the rate at which y is changing:
dy/dt = (3x) / y
dy/dt = (3 * 2) / 6
dy/dt = 1 ft/sec
Therefore, when the firefighter has climbed up 6 feet of the ladder, the top of the ladder is changing at a rate of 1 foot/sec.
Almost forgot
(b) If w is the horizontal distance from the firefighter to the wall, at the instant the angle of the ladder with the ground is π/3, find dw/dt=
Karthik,where is 2 questions??
We draw a diagram with X=distance from wall to base of ladder.
Z=distance along ladder that firefighter has climbed.
H =height of firefighter
W=distance from wall to firefighter
Theta=angle of base of ladder with the ground.
Ladder is sliding away at a constant rate of 3ft/sec------>dx/dt=3
Firefighter is climbing up ladder at rate of 2ft/sec------->dz/dt=2
Cos(theta)=x/29
-sin(theta)d*theta/dt=1/29 dx/dt
Please solve it.....
B)
Cos(theta)=w/(29/-z)
W=(29-z)cos(theta)
dw/dt=(-1 dz/dt)cos(theta)+(29-z)(-sin(theta) d(theta)/dt)
dw/dt=-cos(theta)dz/dt-(29-z)sin(theta)d(theta)/dt
Now you can solve the remaining portion...
Hmm..I don't know whether it is right or not.OK