mr.luna
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?
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Hero
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You got this
Hero
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What's the formula we use here?
mr.luna
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is there a shorcut? i feel like i just have to double something then square it.
Hero
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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
mr.luna
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|dw:1317011985889:dw|
Hero
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I skipped a step which is to find other necessary values
amistre64
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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)
mr.luna
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given: dA/dt= 1 dAREA/dt=2 A=10 AREA = 100
amistre64
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define b as 2A/h
Hero
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Amistre, go to bed, lol
mr.luna
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im gona try and work it out then ill shoot you guys an answer. give me a sec
amistre64
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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?
amistre64
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..... i thought this was bed :)
Hero
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I believe you
mr.luna
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why is b, 2A/h
amistre64
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becasue A = bh/2; Area = base*height/2 ... solveing for base we get; base = 2*Area/height