Question

Consider the curve: y=\frac{ x^4 }{ 16 }+\frac{ 1 }{ 2x^2 } where x in [1,2].
a) Determine the length of the curve.
b) Find the area of the surface obtained by rotating the curve about the x-axis.

Answer 1

\[ y=\frac{ x^4 }{ 16 }+\frac{ 1 }{ 2x^2 }\]
where x \[\in [1,2]\]

Answer 2

Find the derivative of y.
Formula for length: \[l = \int\limits_{x1}^{x2} \sqrt{1+y'^2} dx\]
where y' is the derivative and x1 = 1, x2 = 2.
Hope this helps.

Answer 3

Am I correct in assuming that you're in Calculus II?

There is a formula for finding the length of a given curve from x=a to x=b. Are you familiar with this formula? if so, would you mind looking it up and typing it out?

To help you get started, please find the derivative of \[y=\frac{ x^4 }{ 16 }+\frac{ 1 }{ 2x^2 }\] Square both sides of your result. What next?

Answer 4

Hi @mathmale ~ Yes, Calc II. I have the formula:
\[L=\int\limits_{a}^{b}\sqrt{1+(\frac{ dy }{ dx})^2} dx\]
Then I have the derivative as: \[\frac{ x^3 }{ 4 }-\frac{ 1 }{ x^3 }\]
and squaring both sides I have:
\[\frac{ x^6 }{ 16 }-\frac{ 1 }{ x^6 }\]
It's the next step that I'm not sure of... adding the 1 and taking the square root??

Answer 5

Squaring is wrong !

Answer 6

Let's take a look at your derivative:

\[\frac{ x^3 }{ 4 }-\frac{ 1 }{ x^3 }\]...At first glance it looks fine. Consider using a label:\[\frac{ dy }{ dx}=\frac{ x^3 }{ 4 }-\frac{ 1 }{ x^3 }\]

Answer 7

Next, you'll need to square this derivative. Unfortunately, what you've done isn't kosher.

Note:\[(a-b)^2= a^2 -2ab+b^2, ~NOT~ a^2-b^2.\]

Answer 8

Therefore, \[(\frac{ dy }{ dx })^2 =??\]

Answer 9

ugh! I knew it didn't seem right when I did that...it's the fractions that are throwing me off.... let me redo that....

Answer 10

OK so I have as the squared derivative:
\[\frac{ x^6 }{ 16 }-\frac{ 2x^3 }{4x^3 }+\frac{ 1 }{ x^6}\]

Answer 11

Looks good. Can you simplify the middle term?

Once you've done that, would you add 1 to this expression? (Why do that?)

Answer 12

For reference, where you have \[\frac{ 2x^3 }{ 4x^3 },\] I have 1/5. Why the difference in our results for the middle term?

Answer 13

Like this?
\[\frac{ x^6 }{ 16 }-\frac{ 3 }{ 2 }+\frac{ 1 }{ x^6 }\]

Answer 14

Once we get past this stage, dealing with all those fractions, the rest is often much simpler than you might expect. I'm going to try squaring the derivative once more, to be reasonably certain I have it right. You might want to check your square also.

Answer 15

I get \[(\frac{ dy }{ dx })^2=\frac{ x^6 }{ 16 }-\frac{ 1 }{2 }+\frac{ 1 }{ x^6 }\]

Answer 16

so our disagreement is solely over the middle term. Could you defend your decision to write -3/2 instead of -1/2?

Answer 17

I added the 1 to the middle term of 1/2.

Answer 18

Remember, you're to find the quantity\[1+(\frac{ dy }{ dx })^2\]

Answer 19

and it happens that the middle term of the square of the derivative is -1/2.

What's 1 - (1/2)?

Answer 20

ok....so now that I have that fraction do I take the square root of each term?

Answer 21

Or, taking a wider view, what does \[1+\frac{ x^6 }{ 16 }-\frac{ 1 }{ 2 }+\frac{ 1 }{ x^6 }\] boil down to?

Answer 22

Please let's attend to this algebra first, and then determine what to do with the end result.

Answer 23

\[\frac{ x^6 }{ 16 }-\frac{ 1 }{ 2 }+\frac{ 1 }{ x^6 }\]

Answer 24

that's a POSITIVE 1/2 not a negative...sorry

Answer 25

Better. Now please find the square root of this trinomial. You had a bit of trouble expanding the binomial earlier, so be especially careful as you consider how to find the square root of this latest quantity.

Answer 26

\[\frac{ x^6 }{ 16 }+\frac{ 1 }{ 2 }+\frac{ 1 }{ x^6 }\]

Answer 27

Yes, correct so far. That's a trinomial. Factor it. Find the square root of this trinomial. it's easier to do that after you've factored this trinomial.

Answer 28

Hint: \[a^2 + 2ab +b^2 = (a+b)^2\] is a common "special product."

Answer 29

OK @mathmale - thank you! I now have
\[L=\int\limits_{1}^{2}\frac{ (x^6+4) }{16x^2 }\]

Answer 30

I can't edit that but I have x^6 not squared on the bottom....

Answer 31

I was hoping that you would factor \[\frac{ x^6 }{ 16 }+\frac{ 1 }{ 2 }+\frac{ 1 }{ x^6 }\]

Answer 32

Was I not supposed to find a common denominator and then factor the numerator? I did that and got (x^6+4)^2 in the numerator....

Answer 33

OK...I'm not sure if I'm off track now....but I used
\[\int\limits_{1}^{2}\frac{ x^6+4 }{ 16x^6 }dx \]
and split that into two integrals of \[\int\limits_{1}^{2}\frac{ x^6 }{ 16x^6 }+\int\limits_{1}^{2}\frac{ 4 }{16x^2 }\]
I'm not sure if that's proper though?? to split the fraction like that??
I continued on though and ended up with:
\[L=\frac{ 5 }{ 16 }\ln \left| x^6 \right| \] evaluated at [1,2]
with a final answer of L=1.299650964 ???

Answer 34

Let me respond by asking you where we got such a complicated square in the first place. Note that\[\frac{ dy }{ dx}=\frac{ x^3 }{ 4 }-\frac{ 1 }{ x^3 }\] and that we had to square both sides.

If you experiment, you'll find that \[\frac{ x^6 }{ 16 }+\frac{ 1 }{ 2 }+\frac{ 1 }{ x^6 }\]

Answer 35

Just integrate what lately @mathmale has given you as integrand.

Answer 36

factors to \[(x^3+\frac{ 1 }{ x^3 })^2\]

The square root of that is what?

It's that square root you need to integrate. You do no need to combine those two terms before integration; in fact, it's much easier if you don't combine them. Write 2 integrals instead.

Answer 37

Please write out the integral(s) that you intend to evaluate now. What are the limits of integration? What are the integrands?

Answer 38

ok -
I have \[\int\limits_{1}^{2}\frac{ x^3 }{ 4 }+\frac{ 1 }{ x^3 }\]
integrating I have
\[\frac{ 1 }{ 16 }x^4+\ln \left| x^3 \right|\]
evaluated at [1,2]
with an answer of.... 3.016941542

Answer 39

First term is ok but second term should be -1/x^2

Answer 40

\[\int\limits\limits_{1}^{2}\frac{ 1 }{ x^3 }dx \]does not produce a result that involves logs. Your integrand here is a power function, and therefore you need to apply the power rule for integration.
Hint: Rewrite \[\frac{ 1 }{ x^3 }\]as a negative power of x.