Multivar; Path length, and average y-value over it

[CookieSalesman]

Homework Statement

Letting c(t) =(t²,t,3) for 0<t<1 (Really it's a smaller than or equal to sign)

1. Find the length of the path
2. Find the average y coordinate along the path

Homework Equations

What we're given in the book: (These equations are not necessarily relevant)

(formula 1)

l(c) being the length of the path:

The Attempt at a Solution

So I tried the equation l(c) to find the length.
Pretty sure you integrate

from 0 to 1
The integral gets a bit complicated but, since my professor approved, wolfram alpha told me the arc length of c(t) was approximately 1.4789. I hope everyone has the same answer. Because my book is wrong. (Apparently my textbook is horrendous at being correct, the professor openly tells us)

Could anyone please verify this or tell me why wolfram alpha wouldn't work properly?

But... as for the average y-coordinate...
Isn't it just .5?
By logic you would think that this works out fine.
However my book says either .5736 or .665511 (The book uses log when it really should say ln for some reason, so basically I included both the number for log and ln calculated.
But as for actually trying, I'm unable to figure out what to do.

I can try to apply formula 1, the complicated one above, or I can just try the default single-variable calculus 1/(b-a) int(stuff). That should work, just do y=t and using this you really just integrate this
Since y=t, right?

This just gives 0.5
How in the world am I supposed to use formula 1...?
But also I don't understand why 0.5 isn't correct. I should be able to just ignore the x and z component.

So I recall that formula 1 can be converted to integration by dt
formula 1 =

This would be the formal equation for finding a numerical function over a path c(t).
So I'm converting formula 1 so I can try to apply it.

I don't know why... but I just assumed f(c(t)) could be considered 1. Why... I don't know. Does it make sense if I suppose since the problem asks me to find the... well... yeah I basically have no good reason. I seriously don't get why.
But so now it just looks like

because f(c(t)) is 1. Except ||c'(t)|| is going to only have a y-component. This makes it so I integrate t only.
Applying formula 1...
Now I'm doing 1/ 1.4789
This gives .676178
This seems close enough to the answer .665511.....

But... I seriously have no idea what I did.
I would really appreciate mathematical explanation for part b. And please tell me if my work in part a was proper.

 

[Samy_A]

1) looks correct.

For 2), the assumption $f(c(t))=1$ is wrong, $f(c(t))=t$ for the y coordinate.
So apply the formula, the resulting expression for the average of the y coordinate along the path is $\displaystyle \frac{\int_0^1 t\sqrt{4t²+1}dt}{l(c)}$. This should give you the correct result.Simply assuming that the average of the y coordinate will be 0.5 only works if your path is a straight line.

 

[SteamKing]

However my book says either .5736 or .665511 (The book uses log when it really should say ln for some reason, so basically I included both the number for log and ln calculated.

In calculus, ln and log will be used interchangeably for the logarithm function. Since $\frac{d\, ln(x)}{dx} = \frac{1}{x}$ and $e^{ln\, x} = x$ only apply when speaking of natural logs, this is understood.

The common or base 10 logarithm was used only as a convenience for making numerical calculations with things like slide rules. Except in certain formulas, computers and calculators have cut down on the need to use common logs for calculations anymore.