How do you find an average rate of change in a parabola?
You can start with the typical approach to finding slope for a line: start with points you know satisfy the equation (pick a value of x and solve for y, and do this for 3 points. Using two of these points x1, y1) and (x2, y2), the slope is (y2 - y1)/(x2 - x1). Then do the same calculation for (x2, y2) and (x3, y3). Slope is (y3 - y2)/(x3 - x2). Take these two slopes that you calculated and take the average of the two.
Math question (Quadratic equation of the parabolic arch)?
So I have the arc, total length is 389m high width is 17m Now I'm supposed to use this information to find the parabola's formula. I just don't understand it. We are supposed to use real world examples of parabolas and I picked Tyne's Bridge. Ugh can someone help me? Thank you in advance.
How do you find the area under a parabola?
Find the antiderivative F(x), and evaluate F(b)-F(a), where “a” and “b” are the x values at the left and right ends, respectively, of the area you want to compute:ORJust use a free graphing software:
How do you find the equation of a parabola such that it passes through two given points and the vertex is found on a given line?
Please note: David Seed found a basic mistake in my answer. You should see his answer.This might not be the most straightforward approach. Here's an example, two points and a line. I've deliberately chosen a line with a 'nice' angle. The points are chosen to be a little less 'nice' when they are processed.First rotate the three objects, points and line so that the line corresponds to the x-axis.Now, imagine the vertex of the parabola lying on the [math]x[/math]-axis so that you can model the parabola with the equation, [math]y=k{ \left( x-{ x }_{ 0 } \right) }^{ 2 }[/math], where [math]k[/math] indicates how 'fast' the parabola grows and [math]{ x }_{ 0 }[/math] is the [math]x[/math]-co-ordinate of the vertex of the parabola. You have the co-ordinates of two points on the parabola and you need two parameters in this model. You can therefore calculate them. My model is 0.774974244225257*(x - 0.250951140728542)**2.When I drew this in I got this diagram.To go back to the original co-ordinate system I use the inverse rotation.How you would perform the steps in this agonising process would depend on the tools you have. I'm glad I'm retired.
A bridge has a parabolic arch that is 50 ft high in the center and 150 ft wide @ the bottom..find its height?
Thinking about this, one way to get the answer is to find the parabola that fits the problem. You have an upside down parabola, and, to simplify the work a little, we'll position it so that the highest point is on the y-axis. The formula for the parabola will be y = -ax^2 + c for some values a and c that we have to determine. Three point of the parabola can be determined easily. The highest point is 50 feet, so, since the parabola has its highest point on the y-axis, one point is (0, 50). Using this point, we see that c = 50 in our y = -ax^2 + c equation, since x = 0 cancels out the -ax^2 term. So, our equation so far is y = -ax^2 + 50. To get two more points, use the fact that the bridge is 150 feet wide. So (-75, 0) and (75, 0) are two more points. Using these points and some algebra, we can determine that a = 2/225. So, the parabola that describes the bridge is: y = (-2/225)x^2 + 50 The only thing left to do is plug x = 30 into the equation. If my math is right, the answer is 42 feet.
Prove that the area under a parabolic arch is 2/3 * Base * Height.?
If you use f(x) = h^2 - (ax)^2 then you won't need to use roots in your limits of integration. This is a useful idea in many types of problems. If you want to introduce a constant in algebraic form then introduce it in the form which best suits you.
How do I determine the quadratic equation that passes through two given points?
There are an infinite number of quadratic functions that pass through two points. A quadratic function is defined by three parameters, so it takes three points to pin it down.We can however determine the family of quadratic functions that pass through two points.We have [math]f(x)=ax^2 + bx + c[/math]and two points on [math]f,[/math] [math](k,l)[/math] and [math](m,n).[/math] So[math]l=ak^2+bk+c[/math][math]n=am^2+bm+c[/math]Here [math]a,[/math] [math]b[/math] and [math]c[/math] are our unknowns, and we only have two equations. It’s easiest to eliminate [math]c[/math] by subtracting equations:[math]l-n = a(k^2-m^2) + b(k-m)[/math]So we can see that if [math]k \ne m,[/math] all the quadratic equations that pass through the two points will have parameters [math]a[/math] and [math]b[/math] that lie on a line. We can make [math]a[/math] the independent parameter, and determine [math]b[/math] and [math]c[/math], and thus [math]f,[/math] for each [math]a[/math].[math]b= \dfrac{ l-n - a(k^2-m^2) }{k-m} = \dfrac{l-n}{k-m} - a(k+m)[/math][math]c=l-ak^2-bk[/math]I could substitute in the value for [math]b[/math], but I don’t see that simplifying particularly well, so let’s say you pick [math]a[/math], calculate [math]b[/math], then calculate [math]c[/math].Check:[math]f(k) = ak^2 + bk + c=ak^2+bk + l-ak^2-bk = l \quad \checkmark[/math][math]f(m)=am^2+bm+c[/math][math]=am^2+bm+l-ak^2-bk [/math][math]=a(m^2-k^2)+b(m-k)+l[/math][math]=a(m-k)(m+k)-(k-m)(\dfrac{l-n}{k-m} - a(k+m)) + l[/math][math]=a(m-k)(m+k) - (l-n) + a(k-m)(k+m) + l = n \quad \checkmark[/math]
How do I find out if two points exist on a parabola?
Just to add a little to the existing answers, remember that the numbers you have are of finite precision, therefore it is possible to have numbers that are really, really close to the parabola (or any function), but compute to be on it, or vice versa. Allow a tolerance for how close the result can be before you decide whether it’s on or off the curve.Note that many problems you will be set (in a class) will have answers that ‘work out neatly,’ but that real-world cases are not always so straightforward. There is a long history of fun and games dealing with the issue of whether two lines intersection (see Douglas’s paper “It makes me so CROSS” on the subject).
Quadratic functions / quadratic modelling help?
[AB] is the longest vertical support of a bridge which contains a parabolic arch. There are 15 vertical supports. The vertical supports are 10 m apart. The arch meets the vertical end supports 6 m above the road. a If axes are drawn on the diagram of the bridge above, with x-axis the road and y-axis (AB), find the equation of the parabolic arch in the form y = ax2 + c. b Hence, determine the lengths of all other vertical supports.
Given 3 points in 3D space, how do you find the parabola that passes through them?
The problem is that, even in two dimensions, there isn't really just one parabola that passes through three given points.What is true is that, given three points oriented according to some coordinate system in x and y, all with different x-coordinates, there is a unique parabola passing through those three points that can be described as a function of y in terms of x.So, for example, let's consider the three points [math](-1,1)[/math], [math](0,0)[/math], and [math](2,4)[/math].Obviously one parabola passing through all of these is [math]y=x^2[/math]; but another is [math]x=\frac 12 y^2 - \frac 32 y[/math]:Since there is no canonical way to orient coordinates on your arbitrary plane in 3-space, there is no hope of finding a uniquely specifiable parabola fitting three points.