Question

the altitude of a triangle is increasing at a rate of 1 cm/min while the area in increasing at a rate of 2 cm^2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100cm^2?

Answer 1

You got this

Answer 2

What's the formula we use here?

Answer 3

is there a shorcut? i feel like i just have to double something then square it.

Answer 4

1. Draw diagram, include given info
2. Apply proper formula
3. Take derivative of formula
4. Input values into formula
5. Isolate what you need to find

Answer 5

I skipped a step which is to find other necessary values

Answer 6

h': 1 cm/min

A': 2 cm^2/min.

b'?: h=10 cm, A=100cm^2

A = bh/2

A' = b'h/2 + bh'/2

b' = (A' - bh'/2)(2/h)

Answer 7

given: dA/dt= 1 dAREA/dt=2 A=10 AREA = 100

Answer 8

define b as 2A/h

Answer 9

Amistre, go to bed, lol

Answer 10

im gona try and work it out then ill shoot you guys an answer. give me a sec

Answer 11

b' = (A' - 2Ah'/2h)(2/h)

b' = (A' - Ah'/h)(2/h)

b' = (2 - 100(1)/10) (2/10)

b' = (2 - 10) (1/5)

b' = -8/5 maybe?

Answer 12

..... i thought this was bed :)

Answer 13

I believe you

Answer 14

why is b, 2A/h

Answer 15

becasue A = bh/2; Area = base*height/2 ... solveing for base we get; base = 2*Area/height