The altitude of a triangle is increasing at a rate of 2500 centimeters/minute while the area of the triangle?

Question

increasing at a rate of 1500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10000 centimeters and the area is 86000 square centimeters?

Answer 1

This is a related rate question. We have a relationship between everything through the formula for the area of a triangle:

A = bh/2

If we were to take a derivative with respect to time, we'd have:

dA/dt = (1/2)(b dh/dt + h db/dt)

From the given information, we have an area of 86000 sq cm and an altitude of 10000 cm. We can find the base at this instant in time, again using our area formula:

A = bh/2
86000 = b(10000)/2
86000 = 5000b
b = 17.2 cm

So now, we can go to our derivative and plug in - all that we don't have is db/dt, which is the rate of change of the base:

dA/dt = (1/2)(b dh/dt + h db/dt)
1500 = (1/2)[(17.2)(2500) + (10000)(db/dt)]
3000 = 43000 + (10000)(db/dt)
-40000 = 10000 db/dt
db/dt = -4

So this says that the base is shrinking by 4 cm/min.

Answer 2

Altitude Of A Triangle

Answer 3

section = A = a million/2 * base * top = 0.5bh dA/dt = 0.5 [b * (dh/dt) + h * (db/dt)] If h = 7 cm and A = ninety two cm^2, then.... b = A/0.5h = 2A/h = 184/7 cm dA/dt = 3 cm^2/min dh/dt = 2.5 cm/min (or 5/2) 3 = 0.5 [(184/7) * (5/2) + 7 * (db/dt)] 6 = (460/7) + 7(db/dt) 7(db/dt) = (40 two/7) - (460/7) = - 418/7 cm/min db/dt = (- 418/7)/7 cm/min = - 418/40 9 cm/min ~ - 8.fifty 3 cm/min