The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 98 square centimeters?
A = (1/2)bh
2A = bh
when A = 98 and h = 8
196 = 8b
b = 24.5
2 dA/dt = b dh/dt + h db/dt
subbing in your given data:
dh/dt = 2, dA/dt = 4.5 , b = 24.5, h = 8
2(4.5) = 24.5(2) + 8 db/dt
solve for db/dt
To find the rate at which the base of the triangle is changing, we can apply the chain rule from calculus.
Let's denote the altitude of the triangle as "h" (in cm), the base as "b" (in cm), and the area as "A" (in square cm). We are given that dh/dt = 2 cm/minute (the rate at which the altitude is changing) and dA/dt = 4.5 square cm/minute (the rate at which the area is changing).
We know that the formula for the area of a triangle is A = 1/2 * b * h. We can differentiate this equation implicitly with respect to time (t) to get an equation involving the rates:
dA/dt = (1/2 * b * dh/dt) + (1/2 * h * db/dt)
Since we are interested in finding db/dt (the rate at which the base is changing), we can rearrange the equation and isolate db/dt:
db/dt = (2 * dA/dt - b * dh/dt) / h
Now we can plug in the given values: h = 8 cm, A = 98 square cm, dh/dt = 2 cm/minute, and dA/dt = 4.5 square cm/minute.
db/dt = (2 * 4.5 - b * 2) / 8
To find the value of b, we can use the formula for the area of a triangle:
A = 1/2 * b * h
98 = 1/2 * b * 8
Solving for b, we have:
b = 98 * 2 / 8
b = 24.5 cm
Now we can substitute the known values into the equation for db/dt:
db/dt = (2 * 4.5 - 24.5 * 2) / 8
Simplifying the equation gives us the rate at which the base is changing.