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Marina86 [1]
3 years ago
11

The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Pl

ease don't respond if you can't explain how to get out of an impossible square root. Thanks
Mathematics
1 answer:
Naya [18.7K]3 years ago
7 0
Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
\int_a^b f(r) |r'| dt

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector \sqrt{(x')^2 + (y')^2 + (z')^2}

\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3  dt

\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u}  du \\  \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u}  du \\  \\  \frac{1}{4} \int_0^1 \sqrt{u}  du

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\  \\ =\frac{1}{6} (14^{3/2} - 1)
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Lesechka [4]

First of all, you need to know what a 'solution' is, so you'll know it
when you see it.

A 'solution' to an equation is a number, or a set of numbers, that
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Your equation has two variables in it ... 'x' and 'y' .  In order to find
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The way it stands now, with only one equation, there are actually
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You don't trust me, and you say to me "Wait just a minute there, dude ! 
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You take the first pair and write it into your equation:
x=1, y=0
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This is NOT a true statement.
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I pulled a fast one on you.  If I was charging you money for solutions,
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