The base of a triangle is decreasing at the rate of 1 ft/sec, while the height is increasing at the rate of 2 ft/sec. At what rate is the area of the triangle changing when the base is 10 ft and the height is 70 ft?
a) -25 ft^2/sec
b) -45 ft^2/sec
c) -50 ft^2/sec
d) 45 ft^2/sec
A = 1/2 bh
so,
dA/dt = 1/2 (b dh/dt + h db/dt)
= 1/2 (10 * 2 + 70 * -1)
= -25 ft^2/s
Well, well! Time for some triangle trouble! Let's see what we got here.
We know that the base is decreasing at a rate of 1 ft/sec, so we can say that db/dt = -1 ft/sec. On the other hand, the height is increasing at a rate of 2 ft/sec, so dh/dt = 2 ft/sec.
To find the rate at which the area of the triangle is changing, we can use the formula for the area of a triangle: A = (1/2) * base * height.
Now, we need to find dA/dt, the rate at which the area is changing. To do that, we need to take the derivative of the area equation with respect to time. Let's do that!
dA/dt = (1/2) * (base * dh/dt + height * db/dt)
Plugging in the values we have, we get:
dA/dt = (1/2) * (10 * 2 + 70 * -1) = (1/2) * (20 - 70) = (1/2) * (-50) = -25 ft^2/sec
So, the rate at which the area of the triangle is changing is -25 ft^2/sec.
Looks like option (a) is the right answer! A little negative twist, but that's just how triangles roll sometimes!
To find the rate at which the area of the triangle is changing, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
We are given that the base is decreasing at a rate of 1 ft/sec (db/dt = -1 ft/sec) and the height is increasing at a rate of 2 ft/sec (dh/dt = 2 ft/sec). We want to find the rate at which the area is changing (dA/dt) when the base is 10 ft and the height is 70 ft.
We can start by differentiating the formula for the area with respect to time (t):
dA/dt = (1/2) * (d(base)/dt) * height + (1/2) * base * (d(height)/dt)
We can substitute the given values:
dA/dt = (1/2) * (-1 ft/sec) * 70 ft + (1/2) * 10 ft * (2 ft/sec)
Simplifying the equation:
dA/dt = -35 ft^2/sec + 10 ft^2/sec
dA/dt = -25 ft^2/sec
Therefore, the rate at which the area of the triangle is changing is -25 ft^2/sec.
Therefore, the answer is a) -25 ft^2/sec.
To find the rate at which the area of the triangle is changing, we can use the formula for the area of a triangle: A = (1/2) * base * height.
Given that the base is decreasing at a rate of 1 ft/sec and the height is increasing at a rate of 2 ft/sec, we can express the rates as:
dB/dt = -1 ft/sec (negative since the base is decreasing)
dH/dt = 2 ft/sec (positive since the height is increasing)
To find the rate at which the area is changing, we need to differentiate the area formula with respect to time (t):
dA/dt = (1/2) * (base * dH/dt + height * dB/dt)
Now, substitute the given values:
base = 10 ft
height = 70 ft
dB/dt = -1 ft/sec
dH/dt = 2 ft/sec
dA/dt = (1/2) * (10 * 2 + 70 * -1)
Simplifying further:
dA/dt = (1/2) * (20 - 70)
dA/dt = (1/2) * (-50)
dA/dt = -25 ft^2/sec
Therefore, the rate at which the area of the triangle is changing when the base is 10 ft and the height is 70 ft is -25 ft^2/sec.
Hence, the correct answer is option a) -25 ft^2/sec.