# Math Help - [SOLVED] calc2 major help please!

1. ## [SOLVED] calc2 major help please!

ok so basically we are doing lengths and planes of curves.. i m lost to how the solution manual gets this..

the get:

f '(x) = x (x^2 + 2)^ 1/2

= integral ( 0 to 3) x (x^2 + 2)^ 1/2

im ok till here

int ( 0 to 3) x (1 + {x^2 + 2} x^2)^ 1/2

---------------^---------------^
my question is where did that +1 and x^2 come from how do tthey know they go there?? for other problems how would i know to get those correct values ??

2. Originally Posted by Legendsn3verdie
ok so basically we are doing lengths and planes of curves.. i m lost to how the solution manual gets this..

the get:

f '(x) = x (x^2 + 2)^ 1/2

= integral ( 0 to 3) x (x^2 + 2)^ 1/2

im ok till here

int ( 0 to 3) x (1 + {x^2 + 2} x^2)^ 1/2

---------------^---------------^
my question is where did that +1 and x^2 come from how do tthey know they go there?? for other problems how would i know to get those correct values ??
hmm....maybe I could help if I knew what the original function was.

--Chris

3. Originally Posted by Chris L T521
hmm....maybe I could help if I knew what the original function was.

--Chris
original function is y = (1/3) (x^2+2)^3/2

4. looks like the format for an integral to calculate arc length ...

$S = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$

5. Originally Posted by skeeter
looks like the format for an integral to calculate arc length ...

$S = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$
oh wow so there is a forumal that says add 1 and square the derivative!?

6. Yes. This is the standard formula for arc length.

7. hm ok then why in my original post did the book put a x^2 there instead of just squaring the derivative? instead of squaring the derivatinve the book multiplied the derivative by x^2..

8. Originally Posted by Legendsn3verdie
original function is y = (1/3) (x^2+2)^3/2
Arc length is defined as $L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx$

So we first need to find $f'(x)$, which will then lead to $[f'(x)]^2$

$f'(x)=\frac{1}{2}(x^2+2)^{\frac{1}{2}}\cdot(2x)=x\ sqrt{x^2+2}$

Thus, $[f'(x)]^2=x^2(x^2+2)=x^4+2x^2$

Therefore, the arc length would be $L=\int_0^3\sqrt{1+(x^4+2x^2)}\,dx$

This implies that $L=\int_0^3\sqrt{x^4+2x^2+1}\,dx$

It is good to note that $x^4+2x^2+1$ is a perfect square. So $x^4+2x^2+1=(x^2+1)^2$

Thus, we have $L=\int_0^3\sqrt{(x^2+1)^2}\,dx\implies L=\int_0^3 (x^2+1)\,dx$

Can you take it from here? Does this make sense?

--Chris

9. Originally Posted by Chris L T521
Arc length is defined as $L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx$

So we first need to find $f'(x)$, which will then lead to $[f'(x)]^2$

$f'(x)=\frac{1}{2}(x^2+2)^{\frac{1}{2}}\cdot(2x)=x\ sqrt{x^2+2}$

Thus, $[f'(x)]^2=x^2(x^2+2)=x^4+2x^2$

Therefore, the arc length would be $L=\int_0^3\sqrt{1+(x^4+2x^2)}\,dx$

This implies that $L=\int_0^3\sqrt{x^4+2x^2+1}\,dx$

It is good to note that $x^4+2x^2+1$ is a perfect square. So $x^4+2x^2+1=(x^2+1)^2$

Thus, we have $L=\int_0^3\sqrt{(x^2+1)^2}\,dx\implies L=\int_0^3 (x^2+1)\,dx$

Can you take it from here? Does this make sense?

--Chris
ty sir would kiss ya if icould!

10. Originally Posted by skeeter
looks like the format for an integral to calculate arc length ...

$S = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$

imagine tryin to solve that problem not knowing that forumula! my heart sank threw the floor i was not aware at all! wow that formula is like a light from heaven