1) A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
Find the rate at which the are of the triangle is changing when the base of the ladder is 7 feet from the wall.
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I did the Pythagorean to find the other side of the triangle and got 24. But I'm confused...am I looking for the dr/dt OR dv/dt? Or something else?
Help please!
Rate at which area is changing, dA/dt
where A= 1/2 b*height
dA/dt= 1/2 b dh/dt + 1/2 h db/dt
Wow, thanks for the clarification!
Now I can just plug in b - which is 7 but what about the dh/dt and db/dt? For those unknowns, am I suppose to use the area formula to find it?
No. You are given db/dt
Using the pyth theorm
c^2= b^2 + h^2
dc/dt=0= 2b db/dt + 2h dh/dt
solve for dh/dt in terms of b,h, db/dt
this is what related rates means.
To find the rate at which the area of the triangle is changing, you are looking for dA/dt, which represents the rate of change of the area with respect to time. In this case, we need to find dA/dt when the base of the ladder is 7 feet from the wall.
First, let's draw a diagram to better understand the situation. The ladder, wall, and ground form a right triangle. The ladder (hypotenuse) is 25 feet long, and the base of the ladder is being pulled away from the wall at a rate of 2 feet per second.
Now, let's label the triangle. The base of the ladder can be considered the base of the triangle, and the distance between the base and the wall can be considered the height of the triangle.
Since the ladder is 25 feet long and the base is 7 feet from the wall, we can use the Pythagorean theorem to find the height of the triangle:
c^2 = a^2 + b^2
25^2 = 7^2 + b^2
625 = 49 + b^2
b^2 = 625 - 49
b^2 = 576
b = sqrt(576)
b = 24
So, the height of the triangle is 24 feet.
Now that we have the base and height of the triangle, we can find the area of the triangle using the formula A = (1/2) * base * height:
A = (1/2) * 7 * 24
A = 84 square feet
To find the rate at which the area is changing, we need to differentiate the area function with respect to time (t). Since the base of the ladder is being pulled away from the wall at a rate of 2 feet per second, the rate of change of the base length (b) with respect to time is db/dt = 2 ft/s.
Differentiate the area function A = (1/2) * 7 * 24 with respect to time t:
dA/dt = (1/2) * d(7 * 24)/dt
dA/dt = (1/2) * (7 * db/dt)
dA/dt = (1/2) * (7 * 2)
dA/dt = 7 ft^2/s
Therefore, the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall is 7 square feet per second.