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Find the point on the parametric surface x(s, t) = (s, t^2, t-s) closest to (1/2, 1/4, -1).

Question:

Find the point on the parametric surface {eq}x(s,\ t) = (s,\ t^2, \ t-s) {/eq} closest to {eq}(1/2, \ 1/4 ,\ -1). {/eq}

Parametric Equation:

Parametric equations define an curve as a function of independent parameters.

When parameters are used to define a surface in {eq}\mathbb{R}^{3}{/eq}, it is known as a parametric surface.

Answer and Explanation:

Given

Parametric surface {eq}x\left( {s,t} \right) = \left( {s,{t^2},t - s} \right){/eq} and the point {eq}\left( {1/2,1/4, - 1} \right){/eq}.

Any point {eq}\left( {x,y,z} \right){/eq} on the surface {eq}x\left( {s,t} \right) = \left( {s,{t^2},t - s} \right){/eq} corresponds to {eq}x = s,y = {t^2},z = t - s{/eq}, that is, for {eq}x = s, z = t - x{/eq} or {eq}t = x + z{/eq}, and {eq}y = {\left( {x + z} \right)^2}{/eq}.

Use distance formula to evaluate the distance {eq}D{/eq} between the points {eq}\left( {x,y,z} \right){/eq} and {eq}\left( {1/2,1/4, - 1} \right){/eq}.

{eq}\begin{align*} {\rm{D}} &= \sqrt {{{\left( {x - 1/2} \right)}^2} + {{\left( {y - 1/4} \right)}^2} + {{\left( {z + 1} \right)}^2}} \\ {{\rm{D}}^2} &= {\left( {x - 1/2} \right)^2} + {\left( {y - 1/4} \right)^2} + {\left( {z + 1} \right)^2} \end{align*}{/eq}

Minimize the function {eq}{{\rm{D}}^2}:f = {\left( {x - 1/2} \right)^2} + {\left( {y - 1/4} \right)^2} + {\left( {z + 1} \right)^2}{/eq} subject to {eq}g:{\left( {x + z} \right)^2} - y = 0{/eq} to find the point {eq}\left( {x,y,z} \right){/eq} closest to the point {eq}\left( {1/2,1/4, - 1} \right){/eq} on the surface {eq}x\left( {s,t} \right) = \left( {s,{t^2},t - s} \right){/eq}.

Using Lagrange's Method:

{eq}\begin{align*} \Delta f &= \lambda \Delta g\\ \left( {2\left( {x - 1/2} \right),2\left( {y - 1/4} \right),2\left( {z + 1} \right)} \right) &= \lambda \left( {2\left( {x + z} \right), - 1,2\left( {x + z} \right)} \right) \end{align*}{/eq}

So,

{eq}x - 1/2 = \lambda \left( {x + z} \right),2\left( {y - 1/4} \right) = - \lambda ,z + 1 = \lambda \left( {x + z} \right){/eq}

Solve the above equations.

{eq}\begin{align*} x - 1/2 &= z + 1\\ x &= z + 3/2 \end{align*}{/eq}

{eq}\begin{align*} z + 1 &= \lambda \left( {x + z} \right)\\ z + 1 &= \lambda \left( {z + \dfrac{3}{2} + z} \right)\\ z\left( {1 - 2\lambda } \right) &= \dfrac{{3\lambda - 2}}{2}\\ z &= \dfrac{{3\lambda - 2}}{{2\left( {1 - 2\lambda } \right)}}\\ &= \dfrac{{2 - 3\lambda }}{{4\lambda - 2}} \end{align*}{/eq}

{eq}\begin{align*} x &= z + 3/2\\ &= \dfrac{{2 - 3\lambda }}{{4\lambda - 2}} + \dfrac{3}{2}\\ & = \dfrac{{3\lambda - 1}}{{4\lambda - 2}} \end{align*}{/eq}

{eq}\begin{align*} 2\left( {y - 1/4} \right) &= - \lambda \\ y - \dfrac{1}{4} &= - \dfrac{\lambda }{2}\\ y &= \dfrac{1}{4} - \dfrac{\lambda }{2} \end{align*}{/eq}

Substitute the above values of {eq}\left( {x,y,z} \right){/eq} in the equation {eq}y = {\left( {x + z} \right)^2}{/eq} to evaluate the value of {eq}\lambda {/eq}.

{eq}\begin{align*} y &= {\left( {x + z} \right)^2}\\ \dfrac{1}{4} - \dfrac{\lambda }{2} &= {\left( {\dfrac{{3\lambda - 1}}{{4\lambda - 2}} + \dfrac{{2 - 3\lambda }}{{4\lambda - 2}}} \right)^2}\\ \dfrac{1}{4} - \dfrac{\lambda }{2} &= {\left( {\dfrac{1}{{4\lambda - 2}}} \right)^2}\\ \dfrac{{1 - 2\lambda }}{4} &= \dfrac{1}{{{{\left( {4\lambda - 2} \right)}^2}}}\\ {\left( {4\lambda - 2} \right)^2}\left( {1 - 2\lambda } \right) &= 4 \end{align*}{/eq}

Solving the above equation {eq}{\left( {4\lambda - 2} \right)^2}\left( {1 - 2\lambda } \right) = 4{/eq} for real value of {eq}\lambda {/eq}, such that it satisfies the equation is 0.

Substitute 0 for {eq}\lambda {/eq} to evaluate the point {eq}\left( {x,y,z} \right) = \left( {\dfrac{{3\lambda - 1}}{{4\lambda - 2}},\dfrac{1}{4} - \dfrac{\lambda }{2},\dfrac{{2 - 3\lambda }}{{4\lambda - 2}}} \right){/eq}.

{eq}\begin{align*} \left( {x,y,z} \right) &= \left( {\dfrac{{3\lambda - 1}}{{4\lambda - 2}},\dfrac{1}{4} - \dfrac{\lambda }{2},\dfrac{{2 - 3\lambda }}{{4\lambda - 2}}} \right)\\ &= \left( {\dfrac{{3\left( 0 \right) - 1}}{{4\left( 0 \right) - 2}},\dfrac{1}{4} - \dfrac{0}{2},\dfrac{{2 - 3\left( 0 \right)}}{{4\left( 0 \right) - 2}}} \right)\,\\ &= \left( {\dfrac{1}{2},\dfrac{1}{4}, - 1} \right)\, \end{align*}{/eq}

This implies that the point in question is on the surface, so there exists no point {eq}\left( {x,y,z} \right){/eq} closest to {eq}\left( {1/2,1/4, - 1} \right){/eq} on the surface {eq}x\left( {s,t} \right) = \left( {s,{t^2},t - s} \right){/eq}.


Learn more about this topic:

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Parametric Equations in Applied Contexts

from Precalculus: High School

Chapter 24 / Lesson 6
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