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Let be the curve where .

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maham4102 | eNotes Newbie

Posted September 21, 2013 at 4:39 AM via web

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Let be the curve where .

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steveschoen | College Teacher | (Level 2) Assistant Educator

Posted September 21, 2013 at 10:21 PM (Answer #1)

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Hi, Maham,

We can find this by solving the integral:

`int_a^bsqrt(1+(dy/dx)^2)dx`    from 1 to 4

So, first, we need to find the derivative of y.  That is:

5*x^1.5 --> 7.5 x^0.5

Then, we plug that into the integral.  First, we can simplify under the square root sign:

1 + (7.5x^0.5)^2 = 1 + 56.25x

So, we have:

`int_a^bsqrt(1 + 56.25x)dx`    from 1 to 4

The integral of this is:   (2/168.75)(1+56.25x)^1.5

Plugging in 4 for x, we get:

(2/168.75)(1+56.25*4)^1.5 = 40.27

Then, plugging in 1 for x:

(2/168.75)(1+56.25*1)^1.5 = 5.13

Subtracting these values, the length of the curve is approx.:

40.27 - 5.13 = 35.14 units.

Good luck, Maham.  I hope this helps.

Till Then,

Steve

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