A 13 foot ladder is leaning against the house when it's base starts to slide away. By the time is 12 feet from the house the base is moving at the rate of 5 ft/sec.
1.How fast is the top of the ladder sliding down the wall at that moment?
2. At what rate is the angle theta between the ladder and the ground changing at that moment?
I am unsure of how to find the angle theta.
or #1, use
r^2 = x^2 + y^2
x^2 + y^2 = 169
2x dx/dt + 2y dy/dt = 0 , I differentiated with respect to t (time)
or dy/dt = (-x/y)dx/dt
when x = 12, dx/dt = 5 ft/s
find dy/dt at that moment
12^2 + y^2 = 169
y = 5
dy/dt = (-12/5)(5) ft/s = -12 ft/sec
the negative shows it is descending.
for #2, use
tanØ = y/x
xtanØ = y
xsec^ Ø dØ/dt + tanØ dx/dt = dy/dt
so for our given case:
x = 12 , y = 5, r = 13
tanØ = 5/12 , secØ = 13/12, so sec^2 Ø = 169/144,
dx/dt = +5, dy/dt = -12 , dØ/dt = ????
sub in ...
12(169/144)dØ/dt + (5/12)(5) = -12
I got
dØ/dt = -1
Now remember that the derivatives of the above trig functions are only valid if Ø is in radians, so at that moment, the angle would be decreasing at 1 rad/sec
better check my arithmetic, but it sure came out nicely.
To find the answers to these questions, we can use the concept of related rates. Let's solve each problem step by step.
1. How fast is the top of the ladder sliding down the wall at that moment?
Let's assume that the distance between the base of the ladder and the house is represented by "x" (in feet) and the height from the ground to the top of the ladder is represented by "y" (in feet).
From the given information, we know that dx/dt (rate of change of x) is 5 ft/sec. We need to find dy/dt (rate of change of y) when x = 12 ft.
We can use the Pythagorean theorem to relate x and y:
x^2 + y^2 = 13^2 (by the Pythagorean theorem)
Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0 (since x and y are changing with respect to time)
Now, substitute the values: x = 12 ft (given) and dx/dt = 5 ft/sec (given):
2(12)(5) + 2y(dy/dt) = 0
Simplifying the equation, we get:
24(5) + 2y(dy/dt) = 0
120 + 2y(dy/dt) = 0
2y(dy/dt) = -120
Now, substitute y = √(13^2 - x^2) (from the Pythagorean theorem):
2√(13^2 - x^2)(dy/dt) = -120
Solve for dy/dt:
dy/dt = -120 / [2√(13^2 - x^2)]
Substitute x = 12 ft to find the rate of change of y:
dy/dt = -120 / [2√(13^2 - 12^2)]
Calculate the value to obtain the rate at which the top of the ladder is sliding down the wall.
2. At what rate is the angle theta between the ladder and the ground changing at that moment?
To find the rate at which the angle θ is changing, we can use trigonometry.
By taking the inverse tangent of y/x, we find that tan(θ) = y/x.
Differentiate both sides of this equation with respect to time (t):
[dy/dt * (1/cos^2(θ))] = [dx/dt * (-y/x^2)]
Solve for dθ/dt (rate at which θ is changing):
dθ/dt = [dx/dt * (-y/x^2)] / [1/cos^2(θ)]
Substitute the known values: dx/dt = 5 ft/sec (given), x = 12 ft (given), and dy/dt from step 1:
dθ/dt = [5 * (-y/12^2)] / [1/cos^2(θ)]
Calculate the value to obtain the rate of change of the angle θ at that moment.
Note: You may need to use a calculator or software to calculate the exact values.
To find the rate at which the top of the ladder is sliding down the wall (question 1), and the rate at which the angle theta is changing (question 2), we need to use some basic trigonometry and calculus.
Let's define the variables:
- h: the height of the ladder (distance along the wall)
- x: the distance of the base of the ladder from the house
- L: the length of the ladder (13 feet)
We are given that h = 12 feet and dx/dt (the rate at which x is changing) = 5 ft/sec. We need to find dh/dt (the rate at which h is changing) and d(theta)/dt (the rate at which the angle theta is changing).
Question 1: How fast is the top of the ladder sliding down the wall at that moment?
We know that the ladder, the wall, and the ground form a right-angled triangle. Using the Pythagorean theorem, we can write:
h^2 + x^2 = L^2
Differentiating this equation with respect to time (t), we obtain:
2h * (dh/dt) + 2x * (dx/dt) = 0
Simplifying the equation and substituting the given values, we have:
2 * 12 * (dh/dt) + 2 * 5 * 12 = 0
Solving for dh/dt, we find:
24 * (dh/dt) = -120
dh/dt = -120 / 24 = -5 ft/sec
Therefore, the top of the ladder is sliding down the wall at a rate of 5 ft/sec.
Question 2: At what rate is the angle theta between the ladder and the ground changing at that moment?
The angle theta is defined as the arctan(h/x). To find d(theta)/dt, we need to differentiate this expression with respect to time. Using the quotient rule, we have:
d(theta)/dt = (1 / (1 + (h/x)^2)) * ((1/x) * (dh/dt) - (h/x^2) * (dx/dt))
Substituting the given values, we get:
d(theta)/dt = (1 / (1 + (12/5)^2)) * ((1/5) * (-5) - (12/5^2) * (5))
Simplifying this expression, we find:
d(theta)/dt = (-1/13) * (1 - 12/25) = -13/25 rad/sec
Therefore, the angle theta is changing at a rate of -13/25 rad/sec.
In summary:
1. The top of the ladder is sliding down the wall at a rate of 5 ft/sec.
2. The angle theta between the ladder and the ground is changing at a rate of -13/25 rad/sec.