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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?

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  1. Hero
    • 4 years ago
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    You got this

  2. Hero
    • 4 years ago
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    What's the formula we use here?

  3. mr.luna
    • 4 years ago
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    is there a shorcut? i feel like i just have to double something then square it.

  4. Hero
    • 4 years ago
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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

  5. mr.luna
    • 4 years ago
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    |dw:1317011985889:dw|

  6. Hero
    • 4 years ago
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    I skipped a step which is to find other necessary values

  7. amistre64
    • 4 years ago
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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)

  8. mr.luna
    • 4 years ago
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    given: dA/dt= 1 dAREA/dt=2 A=10 AREA = 100

  9. amistre64
    • 4 years ago
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    define b as 2A/h

  10. Hero
    • 4 years ago
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    Amistre, go to bed, lol

  11. mr.luna
    • 4 years ago
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    im gona try and work it out then ill shoot you guys an answer. give me a sec

  12. amistre64
    • 4 years ago
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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?

  13. amistre64
    • 4 years ago
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    ..... i thought this was bed :)

  14. Hero
    • 4 years ago
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    I believe you

  15. mr.luna
    • 4 years ago
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    why is b, 2A/h

  16. amistre64
    • 4 years ago
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    becasue A = bh/2; Area = base*height/2 ... solveing for base we get; base = 2*Area/height

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