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Curvature Calculator

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Example
Created on 2024-06-20Asked by Layla Hill (Solvelet student)
Find the curvature of the curve y=x2 y = x^2 at the point (1,1) (1, 1) .

Solution

Step 1: Identify the curve y=x2 y = x^2 and the point (1,1) (1, 1) . Step 2: Calculate the first and second derivatives of y y with respect to x x : dydx=2x,d2ydx2=2. \frac{dy}{dx} = 2x, \quad \frac{d^2y}{dx^2} = 2. Step 3: Evaluate the derivatives at the point (1,1) (1, 1) : dydx(1,1)=2(1)=2,d2ydx2(1,1)=2. \frac{dy}{dx}\bigg|_{(1, 1)} = 2(1) = 2, \quad \frac{d^2y}{dx^2}\bigg|_{(1, 1)} = 2. Step 4: Calculate the curvature κ \kappa using the formula: κ=f(x)(1+(f(x))2)32. \kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{\frac{3}{2}}}. Step 5: Substitute the values: κ=2(1+(2)2)32=2(1+4)32=2(5)32=255. \kappa = \frac{|2|}{(1 + (2)^2)^{\frac{3}{2}}} = \frac{2}{(1 + 4)^{\frac{3}{2}}} = \frac{2}{(5)^{\frac{3}{2}}} = \frac{2}{5\sqrt{5}}. Step 6: Conclusion. The curvature of the curve y=x2 y = x^2 at the point (1,1) (1, 1) is 255 \frac{2}{5\sqrt{5}} . Solved on Solvelet with Basic AI Model
Some of the related questions asked by Avery Campbell on Solvelet
1. Calculate the curvature of the curve defined by the parametric equations x=cos(t) x = \cos(t) , y=sin(t) y = \sin(t) , z=t z = t at the point (1,0,π) (1, 0, \pi) 2. Find the radius of curvature of the curve y=x2 y = x^2 at the point (1,1) (1, 1) .,
DefinitionCurvature is the amount by which a geometric object deviates from a straight line in space. It measures the rate of change along the direction of the curve and is defined as the sum of the radius of curvature of a point on the curve. In mathematics, curvature is one of many concepts related to many areas of geometry. This also includes the study of curves and surfaces, such as the development of differential geometry (1868), and the study of motion, such as the study of particle trajectories (1887). It provides detailed information about the geometry and dynamics of moving objects. For example: A circle has non-zero curvature equal to the radius, and a straight line has zero curvature.
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