Use integrals to
a) calculate the circumference of a circle with radius=5
b) calculate the length of the curve
How do I do this?
Use the formula for arc length: the arc length of a continuous function over an interval is given by
A circle with radius can be represented with the equation .
So, you can calculate the circumference of a circle by finding the arc length of the curve from (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, is defined for all real , 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 after which i try to get the derivative of f(x) which is:
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:
and then in the formula:
which is ...
and after a root rule
And this is the hardest part to me. We get:
and after putting in the interval values we get which equals
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 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
which is
When putting this into the formula we get
which is
and putting in the values we get:
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?
So your formula can be derived from mine.
Ok, let's solve it with yours.
Equation of the circle is: .
The circle can't be written as a function of x. But it can be written as two functions.
is the top semi circle and is the bottom semi circle.
Just take one of them and calculate it's length. Let's use the top semi circle, .
Plug it in the formula:
Let , then and the bounds will change too,
Length of the bottom semi circle is . Then circumference of the circle will be .
Nice work! Just one problem:
Your antiderivative is a little off. The general rule is
So, for your integral, we have
Multiplying by two (since we only did the upper-half), we get a circumference of . This agrees with our existing formula, .
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, , so your hypotenuse would be about , 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: where do you get this from? And is the correct answer?
I looked over your work, and it seems good to me. 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 over the interval right? Thus you draw a line through the two endpoints: when , and when , . Finding the distance between these two points we get which is pretty close to the value you found for the length of the actual curve.