Arc Length: y=(4-x^(2/3))^(3/2) from X=1 to 8

[Shutterbug424]

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[mmm4444bot]

Hello Shutterbug:

Do you have any covert confidence regarding your result for f '(x)?

:wink:

The result for f '(x) is not correct.

Your first line looks good, although I'm not sure why you're typing the dx notation. The prime notation is good enough to indicate the derivative of f.

On the second line, I see that you multiplied the factor 3/2 times the factor -2/3 to obtain the simplified factor -1. That's good.

What happened to the square root and its radicand?

(1) The exponent 1/2 seems to have been dropped

(2) The constant 4 changed into 2

(3) The exponent on x somehow changed from 2/3 to 1/3.

You'll need to fix these three errors before continuing. If you're not sure what's wrong, then please explain why you made these changes, and we'll go from there.

Thank you for showing your work.

Also, to answer your question about dropping the radical sign and square at the end: No. We cannot do that because only part of the radicand is being squared.

In other words, you're thinking of doing something like the following.

\(\displaystyle \sqrt{1 - x^2}\)

Drop the square root sign and the exponent 2.

\(\displaystyle 1 - x\)

This is wrong because these two expressions are only equal when x = 0.

The following is true.

\(\displaystyle \sqrt{(1 - x)^2} = |1 - x|\)

Cheers,

~ Mark

 

[Shutterbug424]

I'm not ashamed to admit that many "integral" parts to my Calculus foundation are a little crumbly. :?

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My thought process was that I could somehow distribute through the 1/2 power. I guess I was wrong. Is there a better way to simplify it further before placing it under the radical?

I'll be working on this problem tomorrow as well. For now I'm off to toss and turn and dream of the contents of my daily planner. Thanks for your help thus far. I must admit, you are good...and thorough

 

[mmm4444bot]

I'm not ashamed to admit that many "integral" parts to my Calculus foundation are a little crumbly.

The only cracks in the calculus that I've discussed, so far, are those dx indicators at the end of the derivative. Drop 'em.

The rest of what I talked about has to do with pre-calculus.

Your derivative as stated in the first line is correct; it's the simplifications in lines two and three that are wrong.

You cannot multiply the radicand by the factor 3/2, for two reasons.

(1) You already multiplied it times -2/3. So, it went away. In other words, once you multiply two numbers, you end up with a product. You may not subsequently multiply something else by one of those two numbers without first undoing the first product to get back the individual factors. You're treating the initial derivative result as if there were two factors of 3/2, when there is only one.

(2) Even if you had not first multiplied the factor of 3/2 times -2/3, you still could not multiply it times 4 - x^(2/3) because we may only multiply radicands by other radicands.

As for your attempt to multiply through by 3/2, I don't get it. (3/2)*(4) is not 2. (I'm still not sure how you changed 4 into 2.)

Also, multiplying a power by 3/2 does not affect the exponent. (I'm not sure how the exponent 2/3 changed into 1/3.)

There is no way to simplify the following expression.

\(\displaystyle \left ( 4 - x^{\frac{2}{3}} \right )^{\frac{1}{2}}\)

Therefore, do the simplification of (3/2)*(-2/3) = -1, and you've got f 'x).

OOPS. I just now noticed that your formula for arclength s has a typo. That minus sign should be a plus sign.

\(\displaystyle S \;=\; \int_a^b \sqrt{1 + f '(x)^2} \, dx\)

Can you square your corrected f '(x), and continue?

Cheers,

~ Mark