Problem 15
Question
sketch the curve in the xy-plane. Then, for the given point, find the curvature and the radius of curvature. Finally, $$ y=2 x^{2},(1,2) $$
Step-by-Step Solution
Verified Answer
Curvature is \( \frac{4}{17^{3/2}} \), radius of curvature is \( \frac{17^{3/2}}{4} \), and the curve is a parabola.
1Step 1: Find the First Derivative
The first step to finding curvature is to compute the derivative of the function. For the function \( y = 2x^2 \), the first derivative, which represents the slope of the tangent line, is \( y' = \frac{d}{dx}(2x^2) \). Using the power rule, we get \( y' = 4x \).
2Step 2: Find the Second Derivative
To compute the curvature, we need the second derivative of the function. Differentiate \( y' = 4x \) with respect to \( x \). The second derivative is \( y'' = \frac{d}{dx}(4x) = 4 \).
3Step 3: Use the Curvature Formula
The curvature \( \kappa \) at any point \((x, y)\) for a curve defined by \( y = f(x) \) is given by: \( \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \). At the point \((1,2)\), we have \( y' = 4 \times 1 = 4 \) and \( y'' = 4 \). Substitute these into the curvature formula: \[ \kappa = \frac{|4|}{(1 + 4^2)^{3/2}} = \frac{4}{(1 + 16)^{3/2}} = \frac{4}{17^{3/2}}. \]
4Step 4: Calculate the Radius of Curvature
The radius of curvature \( R \) is the reciprocal of the curvature: \( R = \frac{1}{\kappa} = \left( \frac{17^{3/2}}{4} \right). \)
5Step 5: Sketch the Curve
The equation \( y = 2x^2 \) is a parabola opening upwards. Plot points such as \((0,0)\), \((1,2)\), and \((-1,2)\) to get the shape of the curve. At the point \((1,2)\), the parabola has a specific slope determined by \( y' = 4x \), which helps in visualizing the tangent at that point.
Key Concepts
First DerivativeSecond DerivativeRadius of CurvatureParabola
First Derivative
The first derivative of a function gives us insight into the rate at which the function is changing. In simpler terms, it represents the slope of the tangent line at a given point on the curve. For our exercise involving the function \( y = 2x^2 \), the first derivative \( y' \) is calculated using the power rule for differentiation. The power rule states that if you have a function \( ax^n \), its derivative will be \( anx^{n-1} \).
For \( y = 2x^2 \), applying the rule, we differentiate to get \( y' = 4x \). This means that for every unit change in \( x \), \( y \) changes by \( 4x \). For example, at the point \( x = 1 \), the slope of the tangent line is \( y' = 4 \).
For \( y = 2x^2 \), applying the rule, we differentiate to get \( y' = 4x \). This means that for every unit change in \( x \), \( y \) changes by \( 4x \). For example, at the point \( x = 1 \), the slope of the tangent line is \( y' = 4 \).
- The first derivative provides the gradient of the function at any given point.
- Understanding its value helps in plotting the function accurately.
Second Derivative
The second derivative of a function, denoted by \( y'' \), tells us about the curvature or concavity of the graph of the function. It provides information on how the slope of the tangent line is changing.
For our function \( y = 2x^2 \), once we have the first derivative \( y' = 4x \), we differentiate again to find \( y'' = 4 \). This constant second derivative tells us that the parabola is uniformly curved, with the same rate of change of the slope throughout.
For our function \( y = 2x^2 \), once we have the first derivative \( y' = 4x \), we differentiate again to find \( y'' = 4 \). This constant second derivative tells us that the parabola is uniformly curved, with the same rate of change of the slope throughout.
- A positive second derivative indicates the graph is concave up, like a U-shape.
- Understanding the second derivative helps in determining points of inflection and the nature of curvature.
Radius of Curvature
The radius of curvature is a measure of how sharply a curve bends at a particular point. It is inversely related to the curvature. The formula for the radius of curvature \( R \) is \( R = \frac{1}{\kappa} \), where \( \kappa \) is the curvature.
For our parabola \( y = 2x^2 \) at the point (1, 2), we calculated the curvature \( \kappa \) previously to be \( \frac{4}{17^{3/2}} \). The radius of curvature becomes \( R = \left( \frac{17^{3/2}}{4} \right) \).
For our parabola \( y = 2x^2 \) at the point (1, 2), we calculated the curvature \( \kappa \) previously to be \( \frac{4}{17^{3/2}} \). The radius of curvature becomes \( R = \left( \frac{17^{3/2}}{4} \right) \).
- A larger radius of curvature implies a gentler bend or flatter curve.
- A smaller radius of curvature indicates a sharper curve.
Parabola
A parabola is a symmetric curve shaped like an arch. It is described by a quadratic function of the form \( y = ax^2 + bx + c \). For the equation \( y = 2x^2 \), the parabola opens upwards, indicating that the coefficient of \( x^2 \) is positive. The vertex of this parabola is at the origin (0,0), which is also its minimum point.
- The axis of symmetry for our parabola is along the y-axis.
- The parabola is equidistant from the point (1,2) to its opposite point (-1,2).
Other exercises in this chapter
Problem 14
Complete the squares to find the center and radius of the sphere whose equation is given (see Example 2). \(x^{2}+y^{2}+z^{2}+2 x-6 y-10 z+34=0\)
View solution Problem 15
Sketch the graph of the given cylindrical or spherical equation. $$ r^{2}+z^{2}=9 $$
View solution Problem 15
Find the parametric equations of the line through \((5,-3,4)\) that intersects the \(z\) -axis at a right angle.
View solution Problem 15
Find the equation of the plane through \((2,5,1)\) that is parallel to the plane \(x-y+2 z=4\).
View solution