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In Thomas's calculus $10^{th}$ edition, chapter $5$, exercises $5.3$, problem $9$, I am asked to find the length of an arc with the equation:

$$x=\int^y_0\sqrt{\sec^4(t)-1} dt$$

and $\displaystyle {\frac{-\pi}{4}} \leq y \leq {\frac{\pi}{4}}$.

The problem is, I'm not entirely sure what the equation is supposed to be... The previous problems were relatively straight forward, with simple $x=f(y)$ equations. The second part, the parameters of $y$, is confusing me-- is this a parametric of some sort (considering the "$t$")? I've looked around the chapter and can't seem to find another example like it...

If someone could point me in the right direction, that'd be really helpful.

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The RHS of your equation is of the form f(y), just not explicitly. When you carry out the integral on the RHS you will get [h(t)] evaluated at y and 0. h(t) is the indefinite integral of the integrand here. –  Adam Rubinson May 6 '12 at 23:48

2 Answers 2

up vote 2 down vote accepted

Each value of $y$ gives a particular value of $x$, so that $x$ is a function of $y$. The graph of this function, from $y=-\frac{\pi}{4}$ to $y=\frac{\pi}{4}$ is what you are meant to be computing.

The reason they are giving you this complicated-looking function is that it will make the integral for computing the arc length very easy. Remember that if $x=f(y)$ is a function, then the arc length of the graph of $x=f(y)$ from $y=a$ to $y=b$ is given by $$\mathrm{Length} = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy.$$ If $$x(y) = \int_0^y\sqrt{\sec^4(t)-1}\,dt$$ then by the Fundamental Theorem of Calculus, what is $\frac{dx}{dy}$?

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Thank you very much, you were very helpful. After using the fundamental theorem and then plugging that back into the length formula as $\frac{dx}{dy}$, I ended up with 2, which seems to be what the answer key of the book says as well. Thanks again! –  Andrew M May 6 '12 at 23:59

In general, the length $s$ is given by: $$ s = \int_{t_0}^{t_1} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \, dt $$

In your exercise, $x$ is a function of $y$. Therefore:

$$ s = \int_{y_0}^{y_1} \sqrt{\left(\frac{dx}{dy}\right)^{2} + 1} \, dy $$

Now:

$$ x = \int^y_0 \sqrt{\sec^4(t) - 1} \, dt = F(y) - F(0) $$

Where $F$ is an antiderivative of $\sqrt{\sec^4(t) - 1}$.

Therefore:

$$ \frac{dx}{dy} = F^{\prime}(y) = \sqrt{\sec^4(y) - 1} $$

Can you take it from here?

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Thank you very much for your help! –  Andrew M May 6 '12 at 23:59

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