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OpenStudy (anonymous):

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?

OpenStudy (bahrom7893):

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

OpenStudy (anonymous):

hey

OpenStudy (bahrom7893):

hey will do this one later, i gotta eat

OpenStudy (anonymous):

carra?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

want me to solve it

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Area of triangle = 1/2 base * altitude

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok product rule it is

OpenStudy (anonymous):

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

OpenStudy (anonymous):

b is base, A is area

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (bahrom7893):

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

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