Cornu's Spiral is given by the plane curve \vec{\gamma }\left ( t \right )=\left (...

Question:

Cornu's Spiral is given by the plane curve

{eq}\vec{\gamma }\left ( t \right )=\left ( \int_{0}^{t}\cos \left ( u^2/2 \right )du \right )i+\left ( \int_{0}^{t}\sin \left ( u^2/2 \right )du \right )\hat{j}. {/eq}

Plot this curve and find the curvature, k(s).

Fundamental Theorem of Calculus; Curvature of Planar Curves:

From the Fundamental Theorem of Calculus, we know that if:

{eq}g(t)= \int_{a}^{t} f(u) \; du, {/eq}

then {eq}g'(t)=f(t) {/eq}.

We'll use the following formula for the curvature {eq}k(s) {/eq} of a parameterized planar curve:

{eq}\vec{\gamma }\left ( t \right )=x\left (t \right )\vec{i}+y\left ( t \right )\vec{j}, {/eq}

namely,

{eq}k(s)=k(s(t)) = \frac{{x^{\prime}(t)y^{\prime\prime}(t) - y^{\prime}(t)x^{\prime\prime}(t)}}{{{{\left[ {{{\left( {x^{\prime}(t)} \right)}^2} + {{\left( {y^{\prime}(t)} \right)}^2}} \right]}^{\large\frac{3}{2}}}}}. {/eq}

Answer and Explanation:

The plot of the parameterized curve:

{eq}\begin{align*} \vec{\gamma }\left ( t \right )&=x\left (t \right )\vec{i}+y\left ( t \right )\vec{j}\\ &=\left ( \int_{0}^{t}\cos \left ( u^2/2 \right )du \right )\vec{i}+\left ( \int_{0}^{t}\sin \left ( u^2/2 \right )du \right )\vec{j}, \end{align*} {/eq}

appears in blue in the diagram below, when restricting {eq}t\in[-6,6]. {/eq}

Note: The values of the coordinates of these points were obtained using numerical integration with a CAS system, since their functional expressions cannot be presented in terms of elementary functions.

We know that the curvature {eq}k(s) {/eq} can be evaluated as follows:

{eq}k(s)=k(s(t)) = \frac{{x^{\prime}(t)y^{\prime\prime}(t) - y^{\prime}(t)x^{\prime\prime}(t)}}{{{{\left[ {{{\left( {x^{\prime}(t)} \right)}^2} + {{\left( {y^{\prime}(t)} \right)}^2}} \right]}^{\large\frac{3}{2}}}}}. {/eq}

We'll find the necessary derivatives:

{eq}\begin{align*} x(t)&=\int_{0}^{t}\cos \left ( u^2/2 \right )\,du,\text{ thus by the Fundamental Theorem of Calculus}\\ x^{\prime}(t)=\frac{d\left( x(t) \right)}{dt}&=\cos \left ( t^2/2 \right ),\text{ differentiating again}\\ x^{\prime\prime}(t)&=-t\sin \left ( t^2/2 \right ). \end{align*} {/eq}

Similarly,

{eq}\begin{align*} y(t)&=\int_{0}^{t}\sin \left ( u^2/2 \right )du,\text{ thus by the Fundamental Theorem of Calculus}\\ y^{\prime}(t)=\frac{d\left( y(t) \right)}{dt}&=\sin \left ( t^2/2 \right ),\text{ differentiating again}\\ y^{\prime\prime}(t)&=t\cos \left ( t^2/2 \right ). \end{align*} {/eq}

We know that the arc length parameter {eq}s {/eq} (for {eq}t\ge 0 {/eq}) from the point at {eq}t=0 {/eq} (the origin) is given by:

{eq}\begin{align*} s=s(t)&=\int\limits_{0}^{t} {{{{\left[ {{{\left( {x^{\prime}(u)} \right)}^2} + {{\left( {y^{\prime}(u)} \right)}^2}} \right]}^{\frac{1}{2}}}}}\;du\\ &=\int\limits_{0}^{t} \sqrt{\cos^2 \left ( t^2/2 \right ) + \sin^2 \left ( t^2/2 \right )} \;du\\ &=\int\limits_{0}^{t} \sqrt{1} \;du,\text{ thus}\\ &s=t. \end{align*} {/eq}

Now we can evaluate the curvature {eq}k(s)=k(t) {/eq}:

{eq}\begin{align*} k(s)& = \frac{{x^{\prime}(t)y^{\prime\prime}(t) - y^{\prime}(t)x^{\prime\prime}(t)}}{{{{\left[ {{{\left( {x^{\prime}(t)} \right)}^2} + {{\left( {y^{\prime}(t)} \right)}^2}} \right]}^{\large\frac{3}{2}}}}}\\ & = \frac{{\cos \left ( t^2/2 \right ) \left(t\cos \left ( t^2/2 \right )\right) - \sin \left ( t^2/2 \right ) \left(-t\sin \left ( t^2/2 \right ) \right)}}{1}\\ & = \frac{t{\cos^2 \left ( t^2/2 \right ) +t \sin^2 \left ( t^2/2 \right ) }}{1}\\ & = t\left(\cos^2 \left ( t^2/2 \right ) + \sin^2 \left ( t^2/2 \right )\right),\text{ thus}\\ &\boxed{k(s)=t=s.} \end{align*} {/eq}


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The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10
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