Hello,
Please help!!
Show that the parabola y=ax^2 has its largest curvature at its vertex and no minimum curvature. (Note: since the curvature remains the same if the curve is translated or rotated, this result is true for any parabola)
The curvature of a function is the inverse of the radius of curvature, R
R = [ 1 + (dy/dx)^2 ]^1.5 / d^2y/dx^2
here
R = [ 1+ (2ax)^2 ]^1.5 / 2a
So the curvature, 1/R
=2a/ [1+(2ax)^2]^1.5
now (2ax)^2 is always positive and will be 0 when x = 0
That will give the smallest denominator and therefore the biggest curvature
as x becomes huge (+ or -), the curvature becomes small, but never gets bigger again, so there is no minimum
Hello!
Ah, the parabola, the graceful curve that likes to be the center of attention! Don't worry, I'll do my best to explain the curvature situation.
So, let's talk about curvature. Picture yourself wearing a hula hoop around your waist. The smaller the hula hoop, the more curved it is, right? Well, curvature measures how much a shape, in this case, the parabola, bends at each point.
Now, let's consider the parabola y = ax^2. First things first, we need to find the curvature. And to do that, we need to find the second derivative of the equation.
Taking the derivative two times, we get y'' = 2a. Now, here comes the fun part. Curvature is calculated using the formula: K = |y''| / (1 + (y')^2)^(3/2).
Since y'' = 2a, we can substitute it in the curvature formula and get K = |2a| / (1 + (y')^2)^(3/2).
Now, let's analyze the curvature at different points of the parabola:
1. At the vertex (0, 0): At this point, the slope (y') is zero, so the denominator of the curvature formula becomes 1. Therefore, K = |2a| / 1 = |2a|. The curvature is directly proportional to |2a|.
2. At any other point on the parabola: As we move away from the vertex, the slope (y') increases, and therefore, (y')^2 increases. This means the denominator of the curvature formula becomes larger. Consequently, the curvature decreases.
So, from the analysis above, we see that the curvature is largest at the vertex (0, 0) because the denominator of the curvature formula is minimum there, making the curvature directly proportional to |2a|.
As for the minimum curvature, well, the presence of the absolute value in the formula ensures that the curvature is always positive. Thus, the parabola y = ax^2 doesn't have any minimum curvature.
I hope that brings a curve of amusement to your day! If there's anything else I can help you with, feel free to ask!
To show that the parabola y = ax^2 has its largest curvature at its vertex and no minimum curvature, we need to compute the curvature of the parabola and observe its behavior.
The curvature of a curve can be calculated using the following formula:
Curvature = |(2a)/(1 + (2ax)^2)^(3/2)|
where 'a' represents the constant coefficient, and 'x' is the independent variable.
Let's start by differentiating the equation y = ax^2 with respect to 'x' twice:
dy/dx = 2ax
d^2y/dx^2 = 2a
We obtain the second derivative d^2y/dx^2 = 2a, which is a constant. This tells us that the curvature of the parabola y = ax^2 is constant and does not change with 'x'.
Additionally, notice that the curvature formula does not depend on the value of 'x' or the point on the parabola. It only relies on the coefficient 'a'.
Since the curvature is constant, we can conclude the following:
1. The curvature is the largest at the vertex, where x = 0. Substituting x = 0 into the curvature formula, we get:
Curvature(Vertex) = |(2a)/(1 + (2a*0)^2)^(3/2)| = |(2a)/(1 + 0)^(3/2)|
= |2a/1^(3/2)| = |2a|
Therefore, the curvature at the vertex is given by |2a|, which is the largest possible value for the curvature.
2. The curvature does not have a minimum value since it is constant throughout the entire parabola, irrespective of 'x'.
Hence, we have shown that the parabola y = ax^2 has its largest curvature at its vertex and no minimum curvature.
Sure! To show that the parabola y = ax^2 has its largest curvature at its vertex and no minimum curvature, we need to find the curvature of the parabola and analyze its behavior.
To find the curvature, we'll start by finding the equation of the curve in parametric form. We can parameterize the parabola as x = t and y = at^2.
Next, we find the first and second derivatives of x and y with respect to t:
dx/dt = 1
dy/dt = 2at
d²x/dt² = 0 (second derivative of x is zero since x = t)
d²y/dt² = 2a
Using the formula for curvature (K) in terms of derivatives, we have:
K = |(dy/dt * d²x/dt² - dx/dt * d²y/dt²)| / (dx/dt^2 + dy/dt^2)^(3/2)
Substituting the derivatives we found earlier:
K = |(2at * 0 - 1 * 2a)| / (1^2 + (2at)^2)^(3/2)
K = 2a / (1 + 4a^2t^2)^(3/2)
Now, let's analyze the behavior of the curvature for different values of t.
1. At t = 0 (corresponding to the vertex of the parabola), the curvature is:
K(0) = 2a / (1 + 4a^2 * 0^2)^(3/2) = 2a
So, at the vertex, the curvature is 2a, which is the largest possible curvature for the parabola.
2. As t approaches positive or negative infinity, the denominator (1 + 4a^2t^2)^(3/2) approaches infinity as well. Therefore, the curvature approaches zero.
Based on the above analysis, we can conclude that the parabola y = ax^2 has its largest curvature at the vertex (t = 0) and no minimum curvature elsewhere along the curve.
If you have any further questions, feel free to ask!