A ladder 15 feet long is leaning against a building so that end X is on level ground and end Y is on the wall. Point 0 is where the wall meets the ground. X is moving away from the building at a constant rate of 1/2 foot per second.
(a.) Find the rate in feet per second at which the length of 0Y is changing when X is 9 feet from the building.
(b.) Find the rate of change in square feet per second of the area of triangle X0Y when X is 9 feet from the building
---If you can help answer even part of it, that would be super helpful! Thank you so much :x :x
This is the classic question used by most textbooks to introduce rate of change.
using your definitions
x^2 + y^2 = 15^2
2x dx/dt = 2y dy/dt = 0
x dx/dt + y dy/dt = 0
a) when x = 9
81+y^2 = 225
y =√144 = 12
also given: dx/dt = 1/2
a) 9(1/2) + 12dy/dt = 0
dy/dt = -4.5/12 = -.375
the top of the ladder is sliding down (the negative) at .375 ft/s
b) area = (1/2)xy
d(area)/dt = (1/2)(x dy/dt + y dx/t)
= (1/2)(9(.375) + 12(.5))
= 4.6875
At that moment the area is increasing at aprr. 4.69 ft^2/s
check my arithmetic.
This answer is wrong b/c dy/dt is negative
is part A or part b wrong??
which one is wrong bc the the answer is negative???
Oh, I'd love to help you with this problem! But first, let me warn you that I tend to be more of a clown than a mathematician. So, brace yourself for some silliness!
(a.) To find the rate at which the length of 0Y is changing, we'll need to use the Pythagorean theorem. Let's call the length of 0Y "h" and the distance from X to the building "x". According to the problem, x is changing at a constant rate of 1/2 foot per second. So, we have a right triangle X0Y with legs x (which is decreasing) and h (which is increasing).
Using the Pythagorean theorem, we have x^2 + h^2 = 15^2. To find the rate of change of h, we can differentiate both sides of this equation with respect to time (t).
2x(dx/dt) + 2h(dh/dt) = 0
Now, we're given that dx/dt = -1/2 (since x is decreasing). We're asked to find dh/dt when x = 9. Plugging in these values, we get:
2(9)(-1/2) + 2h(dh/dt) = 0
-9 + 2h(dh/dt) = 0
2h(dh/dt) = 9
dh/dt = 9/(2h)
(b.) To find the rate of change in square feet per second of the area of triangle X0Y, we can use the formula for the area of a triangle:
Area = (1/2)xh
Differentiating both sides of this equation with respect to time (t) gives:
d(Area)/dt = (1/2)(x(dh/dt) + h(dx/dt))
We already know dx/dt and dh/dt from part (a). When x = 9, we have:
d(Area)/dt = (1/2)(9)(9/(2h)) + h(-1/2)
d(Area)/dt = (81/(4h)) - (h/2)
There you have it! I hope my silly explanation was at least somewhat helpful. Good luck with your calculations!
To solve this problem, we can use related rates, which involves finding the rates at which variables are changing with respect to time by using various relationships between the variables involved.
(a.) To find the rate at which the length of OY is changing, we need to determine the relationship between the variables X, OY, and time. We can use the Pythagorean theorem to establish this relationship.
According to the Pythagorean theorem, the ladder length (15 feet) is equal to the square root of the sum of the squares of the horizontal distance (X) and the vertical distance (OY).
The Pythagorean theorem equation is:
15^2 = X^2 + OY^2
Differentiating both sides of the equation with respect to time (t), we get:
0 = 2X * (dX/dt) + 2OY * (dOY/dt)
Since X is moving away from the building, its rate of change (dX/dt) is given as 1/2 foot per second. We need to find the rate of change of OY (dOY/dt) when X is 9 feet from the building.
Plugging in the given values:
0 = 2 * 9 * (1/2) + 2OY * (dOY/dt)
From this equation, we can solve for (dOY/dt), which represents the rate of change of OY.
(b.) To find the rate of change in square feet per second of the area of triangle XOY, we will differentiate the formula for the area of a triangle with respect to time and substitute the given values.
The area of a triangle can be calculated using the formula:
Area = (1/2) * base * height
In this case, the base is X, and the height is OY. So the area formula becomes:
Area = (1/2) * X * OY
We differentiate both sides of the equation with respect to time (t):
(dA/dt) = (1/2) * (dX/dt) * OY + (1/2) * X * (dOY/dt)
Considering that (dX/dt) is given as 1/2 foot per second and we need to find the rate of change of area (dA/dt) when X is 9 feet from the building, we can substitute the given values into the equation to calculate the rate of change of the area.
I hope this helps! Let me know if you have any further questions or need any additional clarification.