Question

Find an equation of a parabola that has curvature 6 at the origin. (Assume the parabola has its vertex at the origin, and opens upward.)

Answer 1

Answer:

The equation of a parabola that has a curvature 6 at the origin is $f(x) = 3\cdot x^{2}$.

Step-by-step explanation:

All parabolas are represented by second-order polynomials, whose standard form is:

$f(x) = a\cdot x^{2}+b\cdot x + c$ (Eq. 1)

Where:

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

$a$, $b$, $c$ - Coefficient of the parabola, dimensionless.

From Vectorial Calculus we know that curvature ($K$), dimensionless, is defined by the following expression:

$K = \frac{\left|\frac{d^{2}y}{dx^{2}} \right|}{\left[1+\left(\frac{dy}{dx} \right)^{2}\right]^{\frac{3}{2} }}$ (Eq. 2)

Where $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$ are the first and second derivatives of the function, respectively. Each expression is described below:

$\frac{dy}{dx} = 2\cdot a\cdot x + b$

$\frac{d^{2}y}{dx^{2}} = 2\cdot a$

If we know that $a > 0$, $b = c = 0$, $x = 0$ and $K = 6$, then the equation of curvature and polynomial are, respectively:

$\frac{2\cdot a}{1^{\frac{3}{2} }} = 6$

$a = 3$

The equation of a parabola that has a curvature 6 at the origin is $f(x) = 3\cdot x^{2}$.

Answer 2

An equation of a parabola that has curvature 6 at the origin is $\rm f(x) = 3x^2$.

We have to determine

An equation of a parabola that has curvature 6 at the origin.

What is the equation of a parabola that has a curvature at the origin?

The equation of a parabola that has a curvature a at the origin is;

$\rm f(x) = ax^2$

The value of a is given by k(0) = 8;

$\rm k(0) = 2a\\\\6 = 2a\\\\a = \dfrac{6}{2}\\\\a=3$

Therefore,

The equation of a parabola that has a curvature a at the origin is;

$\rm f(x) = ax^2$

$\rm f(x) = 3x^2$

Hence, an equation of a parabola that has curvature 6 at the origin is $\rm f(x) = 3x^2$.

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