Question

The altitude of a triangle is increasing at a rate of 2 centimeters/minute while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10 centimeters and the area is 95 square centimeters?

Answer 1

i really think you should practice related rates, working on your other problem right now

Answer 2

hey

Answer 3

hey will do this one later, i gotta eat

Answer 4

carra?

Answer 5

yes

Answer 6

want me to solve it

Answer 7

ok

Answer 8

Area of triangle = 1/2 base * altitude

Answer 9

area is changing with respect to time, base is changing with respect to time and altitude is changing with respect to time . so taking derivative we have

Answer 10

product rule , which is complicated... ok lets look for a relationship given

Answer 11

ok product rule it is

Answer 12

dA/dt = 1/2 b * da/dt + 1/2 db/dt * a , where a is altitude

Answer 13

b is base, A is area

Answer 14

da/dt = 2 "The altitude of a triangle is increasing at a rate of 2 centimeters/minute "
dA/dt = 3.5 "while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute."
At what rate is the base of the triangle changing when the altitude is 10 centimeters and the area is 95 square centimeters?
what is db/dt , when a = 10 , and A= 95

Answer 15

dA/dt = 1/2 b * da/dt + 1/2 db/dt * a
we have to go back and solve for b here to find db/dt

Answer 16

A = 1/2 * b*a, A = 95, a = 10, so b = 19 using algebra.

Answer 17

A = 1/2 * b*a, A = 95, a = 10, so b = 19 using algebra.

Answer 18

plug back into dA/dt = 1/2 b * da/dt + 1/2 db/dt * a
3.5 = 1/2 * 19 *2 + 1/2 * db/dt * 10 , and solve for db/dt

Answer 19

http://i52.tinypic.com/263ww2c.jpg