The altitude of a triangle is increasing at a rate of 3000 centimeters/minute while the area of the triangle is increasing at a rate of 1500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9500 centimeters and the area is 87000 square centimeters?
A = .5 b h
so
87,000 = .5 b (9500)
b = 18.32
then calculus
dA/dt = .5 (b dh/dt + h db/dt)
1500 = .5 ( 18.32 * 3000 + 9500 * db/dt)
strange .....
http://www.jiskha.com/display.cgi?id=1310504428
-13.67?? ...
I got -5.47
If the base were constant the area would be increasing faster than 1500cm^2/min due to the rapid altitude increase. Therefore the base must be decreasing.
positive 5.47
1500 = .5 ( 18.32 * 3000 + 9500 * db/dt)
3000 = 54960 + 9500 db/dt
9500 db/dt = -51,960
db/dt = -5.47
To find the rate at which the base of the triangle is changing, we can use the formula for the area of a triangle:
Area = (1/2) * Base * Altitude
First, let's differentiate both sides of the equation with respect to time (t):
d(Area)/dt = (1/2) * d(Base)/dt * Altitude + (1/2) * Base * d(Altitude)/dt
Given that the rate at which the area of the triangle is increasing (d(Area)/dt) is 1500 square centimeters/minute, and the rate at which the altitude is increasing (d(Altitude)/dt) is 3000 centimeters/minute, we can substitute these values into the equation:
1500 = (1/2) * d(Base)/dt * 9500 + (1/2) * Base * 3000
Since we want to find the rate at which the base is changing (d(Base)/dt), we can rearrange the equation:
d(Base)/dt = (2 * (1500 - (1/2) * Base * 3000)) / (9500)
Now, given that the altitude is 9500 centimeters and the area is 87000 square centimeters, we can substitute these values into the equation to find the rate at which the base is changing:
d(Base)/dt = (2 * (1500 - (1/2) * Base * 3000)) / (9500)
d(Base)/dt = (2 * (1500 - (1/2) * Base * 3000)) / (9500)
d(Base)/dt = (2 * (1500 - 4500 * Base)) / (9500)
Now we can solve for d(Base)/dt.