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How do you use an arc length integral to show the length of the circle of radius r?
Wiki User
∙ 14y ago
The integral from 0 to 2 pi of your constant value r will equal the circumference. This will be equal to 2*pi*r.
This can be derived because of the following:
Arc length = integral from a to b of sqrt(r^2-(dr/dtheta)^2) dtheta. By substituting the equation r = a constant c, dr/dtheta will equal 0, a will equal 0, and b will equal 2pi (the radians in a circle).
By substitution, this becomes the integral from 0 to 2 pi of sqrt(c^2 + 0)dtheta, which leads us back to the original formula.
Wiki User
∙ 14y ago
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What is the geometrical interpretation of line integral?
A line integral or a path or curve integral is a function used when we integrate along a curve. The first step is to parametrize the curve if it is not already in parametric form. So let's look at a general curve, c in its parametric form. The letter t will stand for time. x=h(t) and y=g(t) so both x and y are functions of time. (We assume the curve is smooth) We will integrate over an interval [a,b] so a≤t≤b Now to understand what it means geometrically, let's look at a common geometric shape, the circle. x2 +y2 =16 is the equation of a circle centered at the origin with radius 4. We are going to integrate xy4 along half of the circle. We first need it in parametric form. So x=4cos(t) and y=4sin(t). Keep in mind this circle will be our domain. Just like we find definite integrals over an interval, we use the curve as our analogous domain. We are going to integrate over half the circle so we need to specify t so that our curve will trace out half the circle. We can let -pi/2≤t≤pi/2 Next we compute ds so we need to find dx/dt and dy/dt. So dx/dx=-4sint and dy/dt=4cost. Now we know that ds=Square root (16sin2 t + 16cos2 t)dt=4dt. ( ds is used to show we are moving along the curve) So the integral of xy4 along a general curve c, becomes the integral of cos(t)sin4 (t) dt integrated between -pi/2 to pi/2. The result of integration along this path is 8192/5 so now we ask what does this mean geometrically? (Took a bit of time to get here, but we needed the example) Imagine putting some points on the half circle curve we used. Say point 0, 1, 2, 3, 4 etc. Now draw vectors from 0 to 1, 1 to 2, 2 to 3 etc. We have partitioned our curve into subcurves, the vectors. This is like dividing an interval into the rectangles you have often seen for Riemann integrals and Riemann sums, Instead of looking at the areas of each of those rectangles, we look at the f(pn) where pn are the points 0,1,2,3 etc mentioned above. Just like we add up the sum of the areas of the rectangles for a Riemann sum, we add up the values of the function at each point on our curve, in this case a half circle. Now if we take the limit as the vectors get smaller or as the length of the subintervals along the curve tends toward 0, this is the line integral The path is being partitioned into a polygonal path. We are letting the length or mesh or the partition go to 0. You can do this in 2 or 3 dimension space and we can integrate over a vector field. This last example helps you see an even more geometric example. Sometimes the geometry is hard to see in the examples like the one I gave above. In this one, it is pretty easy. suppose we have a function of two variables that's just a sheet above the x-y plane. Let's look at f(x,y)=5 and we want to take the line integral of this function around the closed path which is the unit circle x2+y2=1. Geometrically, the line integral is the area of a cylinder of length 1 and radius 5. You can check the this is 10Pi ( remember the surface area of a cylinder is 2Pirh, which is 2Pi5x1=10Pi) This one is easier to see geometrically because f(x,y)=5 is a sheet above the x-y plane which is easy to see. If we evaluate the that along the unit circle, then we can see how the cylinder would have to be length or height 1 and radius 5 from the sheet. To tie this to what we said earlier, think of dividing the path, the unit circle, into small subpaths or polygonal paths and evaluating f(x,y)=5 on each of these path. For each one, we get a small piece of the cylinder. If you have trouble seeing this, consider the small piece of the circle between t=0 and t=pi/1000 A very small sector. Now when we look at the value of the function on that very think sector, we have a pie shaped sector with radius 5. We add up lots of these and get the cylinder. Line integrals are also commonly used to find the work done by force long a curve. One last thing to help understand the geometry. Since we talked about vectors and dividing the curve into vectors, let's look at that geometry. Of course, the dot product of any two vectors is positive if the point in the same direction and zero if they are perpendicular. It is negative it they point in roughly opposite directions. The line integral geometrically add up dot products. If F is a vector field and DeltaR is a displacement vector. We look at the dot product of F and DeltaR long the path. If the magnitude of F is constant, the line integral will be a positive number if F is mostly pointing the same way as DeltaR and negative if F is pointing in the opposite direction. If the F is perpendicular to the path at all times, the line integral has a value of zero. We should mention one last thing ( it really is the last thing) Imagine a piece of string or wire. Let that be the curve we have been talking about analogous to the half circle etc. Now suppose we have a function f(x,y) that tells us the mass of the wire or string per unit length. The mass of the string if the line integral of f(x,y) evaluated along the string. An ordinary everyday integral is of y (a function of x) which is just a sum of all the values of the thin rectangles y times dx for all the tiny pieces of dx which make up the x axis. It can be regarded as a line integral along the line known as the x axis. But you could do an integral along any other line or curve, provided you know the value of the function at all points on the line or curve. And that is called a line integral. Instead of the fragment dx, you have a fragment ds where s is your position along the line, just as x describes where you are on the x axis.
Integration of root tanx?
The integral of sqrt(tan(x)) is rather complex and is hard to show with the formatting allowed on Answers.com. See the related links for a representation of the answer.
How is integral calculus used in daily life?
The different aspects of calculus are used in the real world every day. In business, specialists look at the derivatives of trends that can help them predict the future of stocks and markets. Architects commissioned for a job are given a budget and they use optimization to calculate the best amount of material they can get with that budget and space in a building they are designing. The Integral is used to show area under a curve. The indefinite integral is the antiderivative of a function. For these types of professions the integral is their Bible, metaphorically speaking. The watch the trends, convert the data into a quantitative function and then use the integral to predict the future of a company or simply use it with differentiation for an optimization problem. Their are many other uses as well that we use, sometimes subconsciously, in everyday life; these are just a couple of examples.
How do I show that a given function f is continuous everywhere?
Υou show that it is continuous in every element of it's domain.
Show that sin x equals cos x for some angle t between 0 and 90?
sin(x) = cos(x)sin(x)/cos(x) = tan(x) = 1x = arctan(1) = 45 degreessin(45)=cos(45) = Sqrt(2)/2 Answer: By observation. Since Sine = Opposite over hypotenuse and Cosine = Adjacent over hypotenuse. Any right angle triangle where the opposite and adjacent sides are the same length will have Sine equal to Cosine. This only happens with an isosceles triangle (two sides are equal in length). When one angle is 90o the other two are 45o.
How do you work out the circumference of a circle using radius?
The formula for calculating the circumference of a circle is 2πr, where r is the radius of the circle and π is 3.1415926535890793 - usually shorted to either 3.1416 or 3.14 So that the circumference of a circle with a radius of 10 units is 62.83 units There are pi radians in a half of a circle. Thus, the measure of a central angle which is a straight line is pi radians. We have a formula that show that the length of an intercepted arc is equal to the product of the angle in radians that intercepts that arc, with the length of the radius of the circle. So we can say that the length of a semicircle is (pi)(r). In a full circle are 2pi radians. So the length of intercepted arc from a central angle with measure 2pi is 2(pi)(r).
How do you show flow chart for calculating area of circle?
Area of any circle is: pi*radius*radius or pi*r^2
What does a radius in a circle looks like show picture?
It is the line from any point on a circle to the center
What is the circumference of a circle with a radius of 9cm show circulation?
Circumference of the circle: 2*pi*9 = 18*pi cm
What is the area of a circle with the radius of 4 in to the nearest tenth show work?
Area of a circle = (pi) x (radius)2 = (pi) x (4)2 = 16 pi = 50.3 in2 (rounded)
How can you calculate the area of a circle with a radius of 1.4m Please show your method of doing so?
Area of circle = pi*r2 where r is the radius. Area = pi*1.42 = 6.16 m2 (to 2 dp)
How do you show your work to go with how to find the circumference of a circle if the radius is 5?
Circumference = 2*pi*radius In this case radius = 5 units so circumference = 31.4159 units approx.
Show me a 7mm circle?
It is hard to draw one here and do you mean radius 7 mm or diameter 7mm
What is the radius of a circle with the circumference of 13 ft show steps please?
2*pi*radius = circumference 2*pi*radius = 13 Making the radius the subject of the formula is: radius = 13/(2*pi) = 2.06901426 feet Check: 2*pi*2.06901426 = 13 feet
What is the radius of a circle with area 345sq cm and show working out?
Area of a circle = (pi) R2345 = (pi) R2R2 = 345/piR = sqrt(345/pi) = 10.479 cm (rounded)There you are. That's the radius, complete with the working out. You havethe answer, and you've learned nothing, just as you requested.
A pseudocode for finding the area of any circle?
Get radius (as parameter) Calculate area = pi x radius squared Return area The above assumes you write a function or method that calculates the area and returns it. Otherwise: Ask for radius Calculate area = pix radius squared Show area
If a circle has a circumference is 5pi millimeters how do you find the area of the circle show your work also i do not have the radius or any other info?
Circumfernce = 5*pithat is, circumference = pi*d = 5*pi where d is the diameter.Therefore, circumference = 5 millimetres and so radius = d/2 = 2.5 mm.Now that the radius is known, area = pi*r^2 = 19.6 mm^2.
