A ladder 23 feet long leans up against a house.
The bottom of the ladder starts to slip away from the house at 0.16 feet per second.How fast is the tip of the ladder along the side of the house slipping when the ladder is 7.2 feet away from the house? (Round to 3 decimal places.)
ft/sec
Consider the angle the bottom of the ladder makes with the ground.
How fast is the angle changing (in radians) when the ladder is 7.2 feet away from the house?
Well, well, well, looks like we got ourselves a ladder problem. Let's see if we can climb our way to the solution.
First things first, we need to find out how fast the tip of the ladder is slipping along the side of the house. This is basically the rate at which the ladder is moving horizontally.
So, we have a ladder that's 23 feet long and it's slipping away from the house at a rate of 0.16 feet per second. We want to find out how fast the tip of the ladder is moving, let's call that speed "v." We also know that the ladder is 7.2 feet away from the house.
Now, if we imagine a right triangle formed by the ladder, the ground, and the side of the house, we can use some trigonometry to solve this problem. The ladder is the hypotenuse, the side of the house is the adjacent side, and the ground is the opposite side.
Using the Pythagorean theorem, we can say that (side of the house)^2 + (ground)^2 = (ladder)^2. Or, in terms of variables, we have (x)^2 + (h)^2 = (23)^2, where x is the side of the house and h is the height above the ground.
Now, we can differentiate this equation with respect to time to find how the variables are changing over time. We'll call the rate at which x is changing "dx/dt" and the rate at which h is changing "dh/dt" (that's calculus lingo for you).
So, we have 2x(dx/dt) + 2h(dh/dt) = 0, since the ladder length is constant. Rearranging this equation, we get dx/dt = -h(dh/dt)/x.
We want to find dx/dt when x is 7.2 feet away from the house. We also have h = sqrt((23)^2 - (7.2)^2) = sqrt(529 - 51.84) = sqrt(477.16) ≈ 21.848 feet.
Plugging in these values into our equation, we get dx/dt = -21.848(dh/dt)/7.2.
Now, we know that dh/dt is the rate at which the angle is changing, in radians. But hey, who needs radians when you can measure angles in clown noses, am I right?
All kidding aside, we need to find the rate at which the angle is changing in radians when the ladder is 7.2 feet away from the house. This is basically asking for the derivative of the angle with respect to time.
Luckily, we already have an equation that relates the variables x, h, and the angle. It's the good old tangent function: tan(angle) = h/x.
Differentiating this equation implicitly, we get sec^2(angle)(dangle/dt) = (1/x)(dh/dt) - (h/x^2)(dx/dt).
Now, we want to find dangle/dt when x = 7.2 feet. We already know the values for h and dx/dt. Plugging those in, we get sec^2(angle)(dangle/dt) = (1/7.2)(dh/dt) + (21.848/7.2^2)(-21.848/7.2).
Simplifying this a bit, we have sec^2(angle)(dangle/dt) = (1/7.2)(dh/dt) - (21.848/7.2^2)(21.848/7.2).
Now, to find dangle/dt, we just need to divide both sides by sec^2(angle). And guess what, sec^2(angle) is just 1 + tan^2(angle), which equals 1 + (h/x)^2.
So, we have dangle/dt = [(1/7.2)(dh/dt) - (21.848/7.2^2)(21.848/7.2)] / [1 + (h/x)^2].
Now, just plug in the values for dh/dt, h, and x that we found earlier, and you'll have your answer.
But hey, just in case you're wondering, why did the clown bring a ladder to the bar? He heard the drinks were on the house!
Remember to round your answer to 3 decimal places, and happy ladder problem solving!
To find the rate at which the tip of the ladder is slipping along the side of the house, we can use the concepts of related rates.
Let's denote the distance between the bottom of the ladder and the house as x (measured in feet) and the distance between the tip of the ladder and the ground as y (also measured in feet).
We are given that dx/dt = 0.16 ft/s (the rate at which the bottom of the ladder is slipping away from the house) and we need to find dy/dt (the rate at which the tip of the ladder is slipping along the side of the house) when x = 7.2 ft.
Using the Pythagorean theorem, we can relate x and y:
x^2 + y^2 = 23^2
Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Simplifying the equation, we have:
x(dx/dt) + y(dy/dt) = 0
Plugging in the given values (x = 7.2 ft, dx/dt = 0.16 ft/s), we can solve for dy/dt:
(7.2 ft)(0.16 ft/s) + y(dy/dt) = 0
1.152 ft/s + y(dy/dt) = 0
y(dy/dt) = -1.152 ft/s
dy/dt = (-1.152 ft/s) / y
To find y, we can use the Pythagorean theorem:
y = sqrt(23^2 - x^2)
Plugging in the given value (x = 7.2 ft), we have:
y = sqrt(23^2 - 7.2^2)
y ≈ 22.047 ft
Now, we can calculate dy/dt:
dy/dt = (-1.152 ft/s) / 22.047 ft
dy/dt ≈ -0.052 ft/s
Therefore, the tip of the ladder along the side of the house is slipping at a rate of approximately -0.052 ft/s when the ladder is 7.2 feet away from the house.
To find the rate at which the angle is changing, we can use the same equation:
x(dx/dt) + y(dy/dt) = 0
Since the angle can be defined as the inverse tangent of y/x (let's denote it as θ), we can rewrite the equation as:
x(dx/dt) + y(dy/dt) = x(dy/dt) + y(dx/dt) = 0
dy/dt = (-y(dx/dt)) / x
Using the given values (x = 7.2 ft, y ≈ 22.047 ft, dx/dt = 0.16 ft/s), we can calculate dy/dt:
dy/dt = (-22.047 ft)(0.16 ft/s) / 7.2 ft
dy/dt ≈ -0.489 radians/s
Therefore, the angle is changing at a rate of approximately -0.489 radians/s when the ladder is 7.2 feet away from the house.
To solve this problem, we will use the concept of related rates.
Let's denote the length of the ladder as L and the distance between the bottom of the ladder and the house as x. We are given that dL/dt = -0.16 ft/sec and we need to find dx/dt when x = 7.2 feet.
Using the Pythagorean theorem, we can relate L, x, and the height of the ladder (h) as follows:
L^2 = x^2 + h^2
We want to find dx/dt when x = 7.2 feet. To do this, we will differentiate the equation with respect to time (t):
2L(dL/dt) = 2x(dx/dt) + 2h(dh/dt)
Since we are interested in finding dx/dt, we can solve for it:
2L(dL/dt) - 2x(dx/dt) = 2h(dh/dt)
dx/dt = (2L(dL/dt) - 2x(dx/dt)) / 2h
dx/dt = (L(dL/dt) - x(dx/dt)) / h
Now, we need to find h when x = 7.2 feet. Using the Pythagorean theorem, we can find h as follows:
h^2 = L^2 - x^2
h^2 = 23^2 - 7.2^2
h^2 = 529 - 51.84
h^2 = 477.16
h ≈ 21.839 feet
Now we can substitute the given values into the equation for dx/dt:
dx/dt = (23(-0.16) - 7.2(dx/dt)) / 21.839
Let's solve for dx/dt:
dx/dt = (23*(-0.16)) / (21.839 + 7.2)
dx/dt = -3.68 / 29.039
dx/dt ≈ -0.127 ft/sec
So, the tip of the ladder along the side of the house is slipping at a rate of approximately -0.127 ft/sec when the ladder is 7.2 feet away from the house.
distance of foot of ladder from wall ---- x
distance of top of ladder from ground --- y
given: dx/dt = .16 ft/sec
find: dy/dt when x = 7.2 ft
7.2^2 + y^2 = 23^2
y = 21.844
x^2 + y^2 = 23^2
2x dx/dt + 2y dy/dt = 0
we know all the values except dy/dt
2(7.2) + 2(21.844) dy/dt = 0
dy/dt = -7.2/21.844 ft/sec
= appr -0.33 ft/sec
the negative sign tells us the top of the ladder is moving downwards
second part of problem:
cos θ = x/23
x = 23 cosθ
dx/dt = -23sinθ dθ/dt
when x = 7.2
cosθ = 7.2/23
θ = 71.7572..°
sinθ = .94973...
dx/dt = -23sinθ dθ/dt
solve for dθ/dt
explain the negative value of dθ/dt