Calculating arc length of $y=\frac{x^5}{10}+\frac{1}{6x^2}$, $x \in [1, 2]$

48 Views Asked by At

I want to calculate arc length of a function: $y=\frac{x^5}{10}+\frac{1}{6x^2}$, $x \in [1, 2]$. I calculate the derivative of $y$ and multiply it by itself to prepare for the formula:

$((\frac{x^5}{10}+\frac{1}{6x^2})')^2 = \frac{9x^{14}-12x^7+4}{36x^6}$

Now I applied that to the formula below:

$s = \int_1^2\sqrt{1+(f(x)')^2}$

But it produces an integral that I am not able to solve. Are there other ways to go about solving this or how could I possibly calculate that integral?

1

There are 1 best solutions below

2
On BEST ANSWER

This exercise resembles a set of contrived examples commonly used in textbooks when students are studying arc length. For example:

Let $y=\dfrac{x^5}{10}+\dfrac{1}{6x^3}$ with $x^3$ rather than $x^2$ in the second term.

[In these contrived examples, the exponent in the denominator of second term must be two less than the exponent in the numerator of the first term.]

Then one gets $y^\prime=\dfrac{x^4}{2}-\dfrac{1}{2x^4}$.

So $(y^\prime)^2=\dfrac{x^8}{4}-\dfrac{1}{2}+\dfrac{1}{4x^8}$

and $1+(y^\prime)^2=\dfrac{x^8}{4}+\dfrac{1}{2}+\dfrac{1}{4x^8}=\left(\dfrac{x^4}{2}+\dfrac{1}{2x^4}\right)^2$

and one then obtains

$$ s=\int_1^2\dfrac{x^4}{2}+\dfrac{1}{2x^4}\,dx $$

So, is it possible there was a typographical error in the original exercise?