Use integrals to
a) calculate the circumference of a circle with radius=5
b) calculate the length of the curve $\displaystyle y=\frac{2}{3}x^\frac{3}{2}$
How do I do this?
Use the formula for arc length: the arc length $\displaystyle s$ of a continuous function $\displaystyle f$ over an interval $\displaystyle [a,\;b]$ is given by
$\displaystyle s = \int_a^b\sqrt{1+\left(f'(x)\right)^2}\,dx$
A circle with radius $\displaystyle r$ can be represented with the equation $\displaystyle x^2 + y^2 = r^2$.
So, you can calculate the circumference of a circle by finding the arc length of the curve $\displaystyle y = \sqrt{r^2 - x^2}$ from $\displaystyle [-r,\;r]$ (this will give you the top half of it), and then multiplying by 2 to get the whole perimeter.
In order to evaluate the integral, be sure you know your inverse trig rules.
Edit: For your second curve, are you given an interval? If not, $\displaystyle y = \frac23x^{3/2}$ is defined for all real $\displaystyle x\geq0$, and the curve will be infinitely long (as we should expect, the improper integral diverges).
Letīs see now. From the equation of a circle I get $\displaystyle y=\sqrt{25-x^2}$ after which i try to get the derivative of f(x) which is:
$\displaystyle (25-x^2)^\frac{1}{2} = \frac{1}{2}(25-x^2)^{-\frac{1}{2}} (-2x) = \frac{-x}{\sqrt{25-x^2}} $
now Iīll adjust the expression after which Iīll put it in the formula for the perimeter length. Here goes:
Iīll square (is this the correct word for it?) the expression like this:
$\displaystyle \frac{(-x)^2}{25-x^2}$ and then in the formula:
$\displaystyle \int_{-5}^{+5} \sqrt{1+\frac{x^2}{25-x^2}} = \int_{-5}^{+5} \sqrt{\frac{25-x^2}{25-x^2}+\frac{x^2}{25-x^2}}$
which is ...
$\displaystyle \int_{-5}^{+5} \sqrt{\frac{25}{25-x^2}} = \int_{-5}^{+5} \sqrt{\frac{25}{25(1-\frac{x^2}{25})}} = $
$\displaystyle \int_{-5}^{+5} \sqrt{\frac{1}{1-(\frac{x}{5})^2}} $ and after a root rule $\displaystyle \int_{-5}^{+5} \frac{1}{\sqrt{1-(\frac{x}{5})^2}}$
And this is the hardest part to me. We get:
$\displaystyle arcsin\frac{x}{5}\bigg]_{-5}^{+5}$ and after putting in the interval values we get $\displaystyle (arcsin1-arcsin{-1}$ which equals $\displaystyle \pi$
and since we got half the circle we will multiply it with 2 but we still need a 5 to multiply with. Because when I use the easy formula for calculating the perimeter of a circle with radius 5 I get 31,41...
So I have the $\displaystyle \pi$ and I have 2 but where can I get a five (which is the radius in this task)?????
Iīll get back to this one. This Latex thing takes a lot of energy for a newbie. Thanks for your help!
Ok Iīll try. There were interval values, sorry for not posting. They are between 0 and 8 on x-axis.
First Iīll try to differentiate the function like this
$\displaystyle D\frac{2}{3}x^\frac{3}{2} = x^\frac{1}{2}$ which is $\displaystyle \sqrt{x}$
When putting this into the formula we get
$\displaystyle \int_{0}^{8}\sqrt{1+x} = \int_{0}^{8}(1+x)^\frac{1}{2} = \frac{2}{3}(1+x)^\frac{3}{2}\bigg]_0^8 = $
which is
$\displaystyle \frac {2}{3}(1+x)\sqrt{(1+x)}$ and putting in the values we get:
$\displaystyle 18-\frac{2}{3} = 17\frac{1}{3}$ which canīt be true?? Can it? I mean the graph is almost as straigt as a line and when looking at it (sides 8 and 8) the hypotenuse should be about 11,3. But maybe Im wrong?
The arc length formula is actually $\displaystyle \int \sqrt{(dx)^2+(dy)^2}$.
We can calculate it easily by using this formula. Firstly, parameterize the circle.
$\displaystyle x = 5 \cos \theta$
$\displaystyle y=5\sin \theta$
And the integral will be from 0 to 2π.
Find the differentials:
$\displaystyle dx = -5 \sin \theta~d\theta$
$\displaystyle dy = 5 \cos \theta~d\theta$
$\displaystyle \int_0^{2\pi} \sqrt{25 \sin^2 \theta~(d\theta)^2 + 25 \cos^2 \theta (d\theta)^2}$
$\displaystyle \int_0^{2\pi} 5~d\theta = 10\pi$
I read in a book that if we parameterize x itself we get the formula $\displaystyle \int_{-5}^{5}\sqrt{1+f'(x)^2}$
Now I would like to try it with this formula cuz I really donīt understand this parameter stuff. Maybe someday? But there must be a way to calculate the task with the formula. What did I do wrong?
$\displaystyle \int \sqrt{(dx)^2+(dy)^2}$
$\displaystyle \int \sqrt{(dx)^2 \left [1+\left ( \frac{dy}{dx} \right )^2 \right ]}$
$\displaystyle \int \sqrt{1+\left ( \frac{dy}{dx} \right )^2}~dx$
$\displaystyle \int \sqrt{1+ [f'(x)]^2}~dx$
So your formula can be derived from mine.
Ok, let's solve it with yours.
Equation of the circle is: $\displaystyle x^2 + y^2 = 25$.
$\displaystyle y^2 = 25 -x^2$
$\displaystyle y = +\sqrt{25 -x^2}~\text{and}~-\sqrt{25 -x^2}$
The circle can't be written as a function of x. But it can be written as two functions.
$\displaystyle y = \sqrt{25 -x^2}$ is the top semi circle and $\displaystyle -\sqrt{25 -x^2}$ is the bottom semi circle.
Just take one of them and calculate it's length. Let's use the top semi circle, $\displaystyle y = \sqrt{25 -x^2}$.
$\displaystyle f(x) = \sqrt{25 -x^2}$
$\displaystyle f'(x) = \frac{1}{2\sqrt{25 -x^2}}\cdot (25-x^2)'$
$\displaystyle f'(x) = \frac{-2x}{2\sqrt{25 -x^2}}$
$\displaystyle f'(x) = \frac{-x}{\sqrt{25 -x^2}}$
Plug it in the formula:
$\displaystyle \int_{-5}^5 \sqrt{ 1+ \left ( \frac{-x}{\sqrt{25 -x^2}} \right )^2 }~dx$
$\displaystyle \int_{-5}^5 \sqrt{1+ \frac{x^2}{25 -x^2}}~dx$
$\displaystyle \int_{-5}^5 \sqrt{\frac{25}{25 -x^2}}~dx$
Let $\displaystyle x=5u$, then $\displaystyle dx = 5~du$ and the bounds will change too,
$\displaystyle 5\int_{-1}^1 \sqrt{\frac{25}{25 -25u^2}}~du$
$\displaystyle 5\int_{-1}^1 \sqrt{\frac{1}{1 -u^2}}~du$
$\displaystyle 5 \arcsin x \bigg |_{-1}^1$
$\displaystyle 5\pi$
Length of the bottom semi circle is $\displaystyle 5\pi$. Then circumference of the circle will be $\displaystyle 2\cdot 5\pi = 10\pi$.
Nice work! Just one problem:
Your antiderivative is a little off. The general rule is
$\displaystyle \int\frac1{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac xa + C,\;a > 0$
So, for your integral, we have
$\displaystyle \int_{-5}^5\sqrt{\frac{25}{25 - x^2}}\,dx$
$\displaystyle =\int_{-5}^5\frac5{\sqrt{5^2 - x^2}}\,dx$
$\displaystyle =5\int_{-5}^5\frac1{\sqrt{5^2 - x^2}}\,dx$
$\displaystyle =5\arcsin\frac x5\bigg|_{-5}^5 = 5\pi$
Multiplying by two (since we only did the upper-half), we get a circumference of $\displaystyle 2\cdot5\pi = 10\pi$. This agrees with our existing formula, $\displaystyle 2\pi r = 2\pi\cdot5 = 10\pi$.
Regarding the parametric equations: you will probably learn this soon if you haven't already (in the US at least it's usually covered by Calc II). Parametric forms can often be a lot easier to work with.
For your other question:
It's very good that you try to check the reasonableness of your answer like that! However, $\displaystyle \frac23\left(8^{3/2}\right)\approx15.084$, so your hypotenuse would be about $\displaystyle 17.075$, which is pretty close to the arc length that you found (naturally we expect the actual length to be a little larger than this, and it is).
Wow...Iīve never used that arcsin rule before. Great.
Jep, I better do that, it just feels so outer space to me. Too big of a mountain to climb. But thats an obstacle Iīll have to overcome in order to learn new stuff.
I didnīt get this one: $\displaystyle \frac23\left(8^{3/2}\right)\approx15.084$ where do you get this from? And is $\displaystyle 17\frac{1}{3}$ the correct answer?
I looked over your work, and it seems good to me. $\displaystyle 17\frac13 = \frac{52}3$ should be correct.
About your approximation: You used an isosceles triangle with sides of length 8, but this is not correct. You want to approximate the curve $\displaystyle y = \frac23x^{3/2}$ over the interval $\displaystyle [0,\;8],$ right? Thus you draw a line through the two endpoints: when $\displaystyle x = 0$, $\displaystyle \frac23x^{3/2} = 0$ and when $\displaystyle x = 8$, $\displaystyle \frac23x^{3/2} = \frac23\cdot8^{3/2} = \frac23\cdot8\sqrt8 = \frac{32\sqrt2}3\approx15.0849$. Finding the distance between these two points we get $\displaystyle d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}\approx \sqrt{(8 - 0)^2 + (15.0849 - 0)^2}\approx17.075$ which is pretty close to the value you found for the length of the actual curve.