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Magazine Chapter 17: Problem 17
Find the radius of curvature at any point on the given curve. The cycloid \(x=a(t-\sin t), y=a(1-\cos t)\)
Short Answer
Expert verified
The radius of curvature is \(\frac{2 \sqrt{2} a^{1/2} (1 - \cos t)^{3/2}}{a^{3/2} \cos t}\).
Step by step solution
01
Compute the first derivatives
Differentiate the equations for x and y with respect to t, the parameter. \[ \frac{dx}{dt} = a(1 - \cos t) \] \[ \frac{dy}{dt} = a \sin t \]
02
Compute the second derivatives
Differentiate the first derivatives to find the second derivatives. \[ \frac{d^2x}{dt^2} = a \sin t \] \[ \frac{d^2y}{dt^2} = a \cos t \]
03
Use the radius of curvature formula
The formula for the radius of curvature \(R\) for parametric equations \((x(t), y(t))\) is given by: \[ R = \frac{{\bigg( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 \bigg)^{3/2}}}{ \left| \frac{dx}{dt} \frac{d^2y}{dt^2} - \frac{dy}{dt} \frac{d^2x}{dt^2} \right| } \]
04
Substitute the first and second derivatives
Plug the values computed into the radius of curvature formula: \[ R = \frac{ \bigg( a^2 (1 - \cos t)^2 + a^2 \sin^2 t \bigg)^{3/2} }{ \left| a (1 - \cos t) a \cos t - a \sin t a \sin t \right| } \]
05
Simplify the expression under the root
Simplify the numerator: \[ a^2 ((1 - \cos t)^2 + \sin^2 t) = a^2 (2 - 2\cos t) = 2a^2 (1 - \cos t) \] Thus, \[ \big( a^2 (2 - 2\cos t) \big)^{3/2} = (2a^2 (1 - \cos t))^{3/2} = (2a)^{3/2} (1 - \cos t)^{3/2} = 2 \sqrt{2} a^{3/2} (1 - \cos t)^{3/2} \]
06
Simplify the denominator
Simplify the denominator: \[ a^2 ((1 - \cos t) \cos t - \sin^2 t) = a^2 (\cos t - \cos^2 t - \sin^2 t) = a^2 \cos t \]
07
Final calculation of the radius of curvature
Putting it all together, the radius of curvature simplifies to: \[ R = \frac{2 \sqrt{2} a^{3/2} (1 - \cos t)^{3/2}}{ a^2 \cos t } = \frac{2 \sqrt{2} a^{3/2}}{ a^2 } \frac{(1 - \cos t)^{3/2}}{ \cos t } = \frac{2 \sqrt{2} }{ a^{1/2} } \frac{ (1 - \cos t)^{3/2} }{ \cos t } \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
calculus
Calculus is a branch of mathematics that studies continuous change. It has two main branches: differential calculus and integral calculus. Differential calculus is concerned with finding the rate at which quantities change, while integral calculus deals with finding the total size or value, like areas under curves.
In this exercise, we're using differential calculus to find derivatives of parametric equations that describe the cycloid. These derivatives help us understand the rate of change at any given point on the curve, which is crucial for finding the radius of curvature.
In this exercise, we're using differential calculus to find derivatives of parametric equations that describe the cycloid. These derivatives help us understand the rate of change at any given point on the curve, which is crucial for finding the radius of curvature.
cycloid
A cycloid is a type of curve generated by a point on the circumference of a circle as it rolls along a straight line.
The parametric equations for a cycloid are given by:
\(x = a(t - \sin t)\)
\(y = a(1 - \cos t)\)
Here, 'a' is the radius of the generating circle, and 't' is the parameter that represents the angle through which the circle has rotated. Cycloids have interesting properties, one of which is their shape, making them a fascinating subject in both theoretical mathematics and applied physics.
The parametric equations for a cycloid are given by:
\(x = a(t - \sin t)\)
\(y = a(1 - \cos t)\)
Here, 'a' is the radius of the generating circle, and 't' is the parameter that represents the angle through which the circle has rotated. Cycloids have interesting properties, one of which is their shape, making them a fascinating subject in both theoretical mathematics and applied physics.
parametric equations
Parametric equations are a way of defining a curve using a parameter. This is especially useful for complex curves where y cannot be easily expressed as a function of x.
In our example, the cycloid is defined using parametric equations:
\(x = a(t - \sin t)\)
\(y = a(1 - \cos t)\)
Here, both x and y depend on a third variable t (the parameter). This allows us to describe more complex movements and shapes, like the path traced by a rolling circle.
In our example, the cycloid is defined using parametric equations:
\(x = a(t - \sin t)\)
\(y = a(1 - \cos t)\)
Here, both x and y depend on a third variable t (the parameter). This allows us to describe more complex movements and shapes, like the path traced by a rolling circle.
derivatives
Derivatives represent the rate at which a function is changing at any given point and are fundamental in calculus. In the context of this problem, we first find the first derivatives of the parametric equations:
\( \frac{dx}{dt} = a(1 - \cos t) \)
\( \frac{dy}{dt} = a \sin t \)
We then find the second derivatives:
\( \frac{d^2x}{dt^2} = a \sin t \)
\( \frac{d^2y}{dt^2} = a \cos t \)
These derivatives help us understand how the curve bends and turns, which is crucial for our next step: calculating the radius of curvature.
\( \frac{dx}{dt} = a(1 - \cos t) \)
\( \frac{dy}{dt} = a \sin t \)
We then find the second derivatives:
\( \frac{d^2x}{dt^2} = a \sin t \)
\( \frac{d^2y}{dt^2} = a \cos t \)
These derivatives help us understand how the curve bends and turns, which is crucial for our next step: calculating the radius of curvature.
radius of curvature formula
The radius of curvature measures how sharply a curve bends at a particular point. For parametric equations \(x(t), y(t)\), the formula is:
\ R = \frac{\bigg( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 \bigg)^{3/2}}{ \left| \frac{dx}{dt} \frac{d^2y}{dt^2} - \frac{dy}{dt} \frac{d^2x}{dt^2} \right| } \
After substituting the first and second derivatives of our cycloid equations into this formula, and simplifying as shown in the step-by-step solution, we find the radius of curvature at any point on the cycloid. This formula helps us connect the geometric properties of the curve with its analytical representation through calculus.
\ R = \frac{\bigg( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 \bigg)^{3/2}}{ \left| \frac{dx}{dt} \frac{d^2y}{dt^2} - \frac{dy}{dt} \frac{d^2x}{dt^2} \right| } \
After substituting the first and second derivatives of our cycloid equations into this formula, and simplifying as shown in the step-by-step solution, we find the radius of curvature at any point on the cycloid. This formula helps us connect the geometric properties of the curve with its analytical representation through calculus.
