#CALC 2

12 messages · Page 1 of 1 (latest)

strange warren
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well, i guess it is the same way u did part a

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only thing is that if u wanna change the variables name (which u could) u gotta adjust the range for the new variable

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same way u do with integrals

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if 1 <= y <= 2

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u gotta plug those values on f(y) and check ur new interval for x

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and calc the length of f(x)

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im not sure is this is the way tho

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or just integrate normally (?) not sure tho

dawn copper
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y=t
x=t⁴/4 + 1/(8t²)
y'=1
x'=t³ - 1/(4t³)
∫√(1²+(t³-1/(4t³))²)dt = ∫√(1-½+t⁶+1/(16t⁶))dt
= ∫√((t⁶+¼)²/t⁶)dt
= ∫(t⁶+¼)/t³dt

integrated between 1 and 2 is 123/32

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as I'm sure you're aware, you gotta Parameterise your function and apply:
∫√((x'(t))²+(y'(t))²)dt

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@thick rock

thick rock
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thanks