The altitude of a triangle is increasing at a rate of 1500 centimeters/minute while the area of the triangle is increasing at a rate of 1000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10000 centimeters and the area is 99000 square centimeters?
This is a calculus problem.
When the altitude H is 1000 and the area A is 99000, the base width is
B = 2A/B =2*99,000/ 10,000 = 198 cm
A = (1/2)*altitude*base = H*B/2
dA/dt = (1/2)[H*dB/dt + B*dH/dt]
1000 = (1/2)[1000*dB/dt + 198*1500]
solve for dB/dt in cm per minute
1000 = (1/2)[10,000dB/dt + 198*dH/dt]
for dB/dt, i got -295 is that correct?
so i just plug it in to find dH/dt?
To solve this problem, we can use the relationship between the altitude, base, and area of a triangle.
Let's denote the altitude of the triangle as h, the base as b, and the area as A. We are given that dh/dt = 1500 cm/min (change in altitude) and dA/dt = 1000 cm²/min (change in area). We need to find db/dt (the rate at which the base is changing) when h = 10000 cm and A = 99000 cm².
The formula for the area of a triangle is A = (1/2) * b * h.
Differentiating both sides with respect to time (t), we have:
dA/dt = (1/2) * (db/dt * h + b * dh/dt)
Substituting the given values, we have:
1000 = (1/2) * (db/dt * 10000 + b * 1500)
Now, let's rearrange the equation to solve for db/dt:
db/dt = (2000 - 1500b) / 10000
To find db/dt when h = 10000 cm and A = 99000 cm², we substitute these values into the equation:
db/dt = (2000 - 1500b) / 10000
Since we're given A = 99000 cm², we can use the formula for the area of a triangle to find the value of b:
99000 = (1/2) * b * 10000
Simplifying, we find:
b = 19.8 cm
Substituting this value of b into the equation for db/dt, we get:
db/dt = (2000 - 1500 * 19.8) / 10000
db/dt = -4017 / 10000
So, the rate at which the base of the triangle is changing when the altitude is 10000 cm and the area is 99000 cm² is approximately -0.4017 cm/min.