"How fast is the area of the triangle formed by the ladder, the house, and the ground when the base of the ladder is 7 ft. away from the wall?
Given
h=24ft.
dh/dt=-7/12ft. per sec.
b=7 (base)
db/dt=2ft. per sec.
d/dt[A=1/2bh]
How would you derive this with respect to time?
A= 1/2 bh
dA/dt= 1/2 hdh/dt + 1/2 bdb/dt
compute, you have h, dh/dt, b, db/dt
thanks
To find the rate of change of the area of the triangle with respect to time, you can use the chain rule of differentiation.
The formula for the area of a triangle is A = 1/2 * base * height.
In this case, the base (b) is given as 7 ft., and its rate of change with respect to time (db/dt) is given as 2 ft. per sec.
The height (h) is given as 24 ft., and its rate of change with respect to time (dh/dt) is given as -7/12 ft. per sec.
Now, let's differentiate the area A with respect to time (t) using the chain rule.
d/dt[A] = d/dt[(1/2) * b * h]
Using the product rule, we can differentiate this function step by step:
1. Differentiating (1/2) * b * h with respect to t:
d/dt[(1/2) * b * h] = (1/2) * (db/dt) * h + (1/2) * b * (dh/dt)
2. Now substitute the given values:
d/dt[A] = (1/2) * (2 ft. per sec.) * 24 ft. + (1/2) * 7 ft. * (-7/12 ft. per sec.)
3. Simplify the expression:
d/dt[A] = 24 ft./sec. + (-49/24 ft./sec.)
d/dt[A] = (-1/24) ft./sec.
Therefore, the rate of change of the area of the triangle with respect to time is -1/24 ft./sec.
To find how fast the area of the triangle is changing with respect to time, we need to use the formula for the area of a triangle, A = 1/2 * base * height.
Given:
h = 24 ft (height)
dh/dt = -7/12 ft per second (rate of change of height)
b = 7 ft (base)
db/dt = 2 ft per second (rate of change of base)
To derive the rate of change of the area with respect to time, we differentiate the formula for the area of a triangle, A = 1/2 * b * h, with respect to time.
d/dt[A] = d/dt[1/2 * b * h]
Using the product rule of differentiation, we can derive the equation as follows:
d/dt[A] = 1/2 * (d/dt[b * h]) + (b * h) * (d/dt[1/2])
The derivative of the first term, d/dt[b * h], involves the rates of change of both the base and the height:
d/dt[b * h] = db/dt * h + b * dh/dt
Since h = 24 ft and dh/dt = -7/12 ft per second, we can substitute these values into the equation:
d/dt[b * h] = db/dt * h + b * dh/dt
= (2 ft/s) * (24 ft) + (7/12 ft/s) * (7 ft)
= 48 ft^2/s + 49/12 ft^2/s
Now, let's substitute this value back into the original equation:
d/dt[A] = 1/2 * (d/dt[b * h]) + (b * h) * (d/dt[1/2])
= 1/2 * (48 ft^2/s + 49/12 ft^2/s) + (7 ft) * (24 ft) * (0)
= 1/2 * (48 ft^2/s + 49/12 ft^2/s)
= (48/2) ft^2/s + 49/24 ft^2/s
= 24 ft^2/s + 2.04 ft^2/s
Therefore, the rate at which the area of the triangle is changing with respect to time is 24 ft^2/s + 2.04 ft^2/s, or approximately 26.04 ft^2/s.