FP2 Arc Length

A curve has the equation

y=sinh^{-1}x + x \sqrt {1+x^{2}}

. The length of the arc of the curve between the points where

x=0

and

x=1

is denoted by

L

.

Show that

L = \frac {5ln(5)+12}{8}

---

My working:

\frac {dy}{dx} = 2 \sqrt{1+x^{2}}

L = \int^1_0 \sqrt {1+ (\frac {dy}{dx})^{2}}\ dx

L = \int^1_0 \sqrt {1+4(1+x^{2})}\ dx

L = \int^1_0 \sqrt {4x^{2}+5}\ dx

I then used the substitution

x=\frac {\sqrt {5}}{2}tan \theta

which led me to this:

L = \frac{5}{2} \int^{tan^{-1} (\frac{2}{\sqrt{5}})}_0 sec^{3}\theta\ d\theta

Is what I have done so far correct?

(edited 10 years ago)

Reply 1

brianeverit

Original post by SherlockHolmes

A curve has the equation

y=sinh^{-1}x + x \sqrt {1+x^{2}}

. The length of the arc of the curve between the points where

x=0

and

x=1

is denoted by

L

.

Show that

L = \frac {5ln(5)+12}{8}

---

My working:

\frac {dy}{dx} = 2 \sqrt{1+x^{2}}

L = \int^1_0 \sqrt {1+ (\frac {dy}{dx})^{2}}\ dx

L = \int^1_0 \sqrt {1+4(1+x^{2})}\ dx

L = \int^1_0 \sqrt {4x^{2}+5}\ dx

I then used the substitution

x=\frac {\sqrt {5}}{2}tan \theta

which led me to this:

L = \int^{tan^{-1} (\frac{2}{\sqrt{5}})}_0 sec^{3}\theta\ d\theta

Is what I have done so far correct?

I think a hyperbolic substitution is better.

Reply 2

Jooooshy

Original post by SherlockHolmes

A curve has the equation

y=sinh^{-1}x + x \sqrt {1+x^{2}}

. The length of the arc of the curve between the points where

x=0

and

x=1

is denoted by

L

.

Show that

L = \frac {5ln(5)+12}{8}

---

My working:

\frac {dy}{dx} = 2 \sqrt{1+x^{2}}

L = \int^1_0 \sqrt {1+ (\frac {dy}{dx})^{2}}\ dx

L = \int^1_0 \sqrt {1+4(1+x^{2})}\ dx

L = \int^1_0 \sqrt {4x^{2}+5}\ dx

I then used the substitution

x=\frac {\sqrt {5}}{2}tan \theta

which led me to this:

L = \int^{tan^{-1} (\frac{2}{\sqrt{5}})}_0 sec^{3}\theta\ d\theta

Is what I have done so far correct?

You're missing a constant before your integral..

Reply 3

SherlockHolmes

Original post by brianeverit

How on earth did you get that result for dy/dx?

y=sinh^{-1}x + x \sqrt {1+x^{2}}

\frac {dy}{dx} = \frac {1}{\sqrt{1+x^{2}}} + x \cdot \frac {d(\sqrt{1+x^{2}})}{dx} + \sqrt{1+x^{2}} \cdot 1

\frac {dy}{dx} = \frac {1}{\sqrt{1+x^{2}}} + x^{2}(1+x^{2})^{\frac{1}{2}} + \sqrt{1+x^{2}}

\frac {dy}{dx} = \frac {1}{\sqrt{1+x^{2}}} + \frac {x^{2}}{\sqrt{1+x^{2}}} + \sqrt{1+x^{2}}

\frac {dy}{dx} = \frac {2(1+x^{2})}{\sqrt{1+x^{2}}}

\frac {dy}{dx} = \frac {2(1+x^{2}) \sqrt{1+x^{2}}}{1+x^{2}}

\frac {dy}{dx} = 2\sqrt{1+x^{2}}

Reply 4

SherlockHolmes

Original post by Jooooshy

You're missing a constant before your integral..

I wrote it down but forgot to type it in...

Thanks.

Original post by brianeverit

I think a hyperbolic substitution is better.

x=\frac{\sqrt{5}}{2}sinh\theta

?

EDIT: The substitution above worked. Thanks.

(edited 10 years ago)

Reply 5

brianeverit

Original post by SherlockHolmes

I wrote it down but forgot to type it in...

Thanks.

x=\frac{\sqrt{5}}{2}sinh\theta

?

EDIT: The substitution above worked. Thanks.

Your welcome. Sorry about the previous post. I realised what you had done as soon as I had sent it but couldn't delete it.

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