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
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
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.