The altitude of a triangle is increasing at a rate of 1.000 centimeters/minute while the area of the triangle is increasing at a rate of 3.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11.500 centimeters and the area is 85.000 square centimeters?
3 answers
Note: The "altitude" is the "height" of the triangle in the formula "Area=(1/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shape the triangle really is.
(1/2) b h = A
(1/2) b (11.5) = 85
b = 14.8 if we need it
2 dA/dt = b dh/dt + h db/dt
2 (3) = 14.8 (1) + 11.5 db/dt
- 8.8 = 11.5 db/dt
db/dt = - .765
(1/2) b (11.5) = 85
b = 14.8 if we need it
2 dA/dt = b dh/dt + h db/dt
2 (3) = 14.8 (1) + 11.5 db/dt
- 8.8 = 11.5 db/dt
db/dt = - .765
Water is leaking out of an inverted conical tank at a rate of 10600.0 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 m and the the diameter at the top is 6.0 m. If the water level is rising at a rate of 27.0 cm/min when the height of the water is 1.5 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute.