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Engineering Calculus Notes 320

# Engineering Calculus Notes 320 - any curve in S −→ γ t...

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308 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION the parameters constant will give a vector in this plane: holding s constant at s = s 0 , we can take r = r 0 + t to get −→ γ ( t ) = −→ p ( r 0 + t,s 0 ) whose velocity at t = t 0 is −→ v r (0) = −→ p ∂r and similarly, the velocity obtained by holding r = r 0 and letting s = s 0 + t will be −→ v s (0) = −→ p ∂s . Because P is a regular point, these are linearly independent and so form direction vectors for a parametrization of a plane T ( r 0 ,s 0 ) −→ p ( r 0 + r,s 0 + s ) = −→ p ( r 0 ,s 0 ) + r −→ p ∂r + s −→ p ∂s . By looking at the components of this vector equation, we easily see that each component of T ( r 0 ,s 0 ) −→ p ( r 0 + r,s 0 + s ) is the linearization of the corresponding component of −→ p ( r,s ), and so has first order contact with it at t = 0. It follows, from arguments that are by now familiar, that for any curve in S −→ γ
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Unformatted text preview: any curve in S −→ γ ( t ) = −→ p ( r ( t ) ,s ( t )) = ( x ( r ( t ) ,s ( t )) ,y ( r ( t ) ,s ( t )) ,z ( r ( t ) ,s ( t ))) the velocity vector −→ v (0) = ∂ −→ p ∂r dr dt + ∂ −→ p ∂s ds dt lies in the plane parametrized by T −→ p . It is also a straightforward argument to show that this parametrization of the tangent plane has frst order contact with −→ p ( r,s ) at ( r,s ) = ( r ,s ), in the sense that v v v −→ p ( r + △ r,s + △ s ) − T ( r ,s ) −→ p ( r + △ r,s + △ s ) v v v = o ( b ( △ r, △ s ) b ) as ( △ r, △ s ) → −→ ....
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