## mr.luna 4 years ago 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?

1. Hero

You got this

2. Hero

What's the formula we use here?

3. mr.luna

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

4. Hero

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

5. mr.luna

|dw:1317011985889:dw|

6. Hero

I skipped a step which is to find other necessary values

7. amistre64

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)

8. mr.luna

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

9. amistre64

define b as 2A/h

10. Hero

Amistre, go to bed, lol

11. mr.luna

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

12. amistre64

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?

13. amistre64

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

14. Hero

I believe you

15. mr.luna

why is b, 2A/h

16. amistre64

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