Homework 4 Solutions

Homework 4 Solutions - WEEK 4 HOMEWORK SOLUTIONS 1 a. Our...

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Unformatted text preview: WEEK 4 HOMEWORK SOLUTIONS 1 a. Our curve is parameterized by r(t) = (x(t), y (t)) for t ∈ R. To calculate curvature using the formula (1) κ( t) = | v ( t) × a ( t) | , | v ( t) | 3 we must first write our parameterization as a three dimensional vector, then compute the associated velocity and acceleration. We can stick the plane that this curve is stuck in at any place along the z-axis, meaning we should just set the z component equal to the constant c. We then get; r ( t ) = ( x ( t ) , y ( t ) , c) v(t) = (x￿ (t), y ￿ (t), 0) a(t) = (x￿￿ (t), y ￿￿ (t), 0). To compute v (t) × a(t) it is best to first recognize that the only nonzero component ˆ will be k . We find that ˆ v(t) × a(t) = (x￿ (t)y ￿￿ (t) − x￿￿ (t)y ￿ (t))k |v(t) × a(t)| = |x￿ (t)y ￿￿ (t) − x￿￿ (t)y ￿ (t)|. Input this into 1 we get κ( t) = b. Define the function K (t) as K ( t) = N(t0 ) · (r(t0 ) − r(t)) . | r( t0 ) − r( t ) | 2 |x￿ (t)y ￿￿ (t) − x￿￿ (t)y ￿ (t)| , (x￿ (t)2 + y ￿ (t)2 )( 3/2) If we let θ be the angle between the two vectors in the denomiator we may writ this expression in terms of the respective magnitudes and θ. Writing this out and using 1 2 WEEK 4 HOMEWORK SOLUTIONS the fact that N(t) is a unit vector we get that K ( t) = |N(t0 )||r(t0 ) − r(t)| cos(θ) | r( t0 ) − r( t ) | 2 cos(θ) = | r( t0 ) − r( t ) | c. Now, we have oriented our coordinate system and set our initial time such that t0 = 0, r(0) = (0, 0), and N(0) = (0, 1). Since we are entirely in the x-y plane and have oriented our coordinate system so that N(0) = (0, 1) we must have that T(0) = (1, 0). So, for at least small t, our motion is completely in the x direction. Since t is arclength, the magnitudes should be approximately the same as well. This can also be seen by doing a first order Taylor expansion on the x coordinate around small t and using that x(0) = 0 and x￿ (0) = 1. The second order Taylor expansion for the y coordinate is given by (2) y (t) ≈ y (0) + y ￿ (0)t + y ￿￿ (0) 2 t 2 Since we are at the origin when t = 0 we have that y (0) = 0. The y coordinate of T(0) gives us y ￿ (0) = 0, so the Taylor expansion reduces to y ￿￿ (0) 2 t. 2 Using the approximations we have just justified we can simplify our expression for κ(t) by using: (3) y ( t) ≈ x￿ ( t ) x￿￿ (t) y ￿ ( t) y ￿￿ (t) Putting this into 1 we find that κ( t) ≈ |y ￿￿ (0) − 0 · y ￿￿ (0)t| |y ￿￿ (0)| = (12 + (y ￿￿ (0)t)2 )3/2 (1 + (y ￿￿ (0)t)2 )3/2 ≈ ≈ ≈ ≈ 1 0 y ￿￿ (0)t y ￿￿ (0) d. We want to find an expression for cos(θ) using our approximations. We have that N(0) = (0, 1) and for small t: r(t) = (t, y ￿￿ (0) 2 t) 2 WEEK 4 HOMEWORK SOLUTIONS 3 Taking the dot product we get that cos(θ) ≈ = = N(0) · r(0) |N(0)||r(0)| (0, 1) · (t, y |(t, ￿￿ (0) y (0)t ￿￿ (0) 2 y ￿￿ (0) 2 t )| 2 ￿￿ 2 2 t2 ) 2(t2 + ( y t 2 ) 2 ) 1/ 2 e. Now, we know that |r(t)| is given by (t2 +( y for cos that we derived above, by r(t) we get ￿￿ (0) 2 t2 )2 )1/2 . So, dividing our expression cos(θ) y ￿￿ (0)t2 = ￿￿ (0) r( t) 2(t2 + ( y 2 t2 )2 ) y ￿￿ (0) = ￿￿ 2 2(1 + y (0) t2 ) 4 So, taking the limit we find that lim cos(θ) y ￿￿ (0) = lim ￿￿ 2 t→0 r( t ) t→0 2(1 + y (0) t2 ) 4 y ￿￿ (0) = 2 κ(0) = 2 Actually, the last equality only holds if y ￿￿ (0) ≥ 0. Fortunately, near the origin the y component is approximately a function of x and since N(0) = (0, 1), y (x) must be concave up. Meaning, y ￿￿ (0) ≥ 0. From this we can see that K (t) will be continuous at t = 0 if κ(0) < ∞. 2 Consider the function f (x, y ) = x2 − y 2 . x2 + y 2 4 WEEK 4 HOMEWORK SOLUTIONS From the level curves it is easy to see that the limit of the function at the origin depends on the direction that we approach it from. So, this function can not be continuous at the origin. So, the function is continuous on R \ {0}. Now, we are asked to show that lim f (x, y ) = 0. (x,y )→(1,1) We can transform this function into polar coordinates to get (r cos(θ))2 − (r sin(θ))2 (r cos(θ))2 + (r sin(θ))2 r2 (cos2 (θ) − sin2 (θ)) r2 (cos2 (θ) + sin2 (θ)) cos2 (θ) − sin2 (θ) cos(2θ) f (r, θ) = = = = WEEK 4 HOMEWORK SOLUTIONS 5 So, this function doesn’t depend on the distance from the origin at all. To evaluate this limit we must also figure out the coordinates of the point we are approaching. √ √ r= 12 + 12 = 2 π θ = arctan(1) = 4 So, now we are trying to show that θ→ π 4 lim cos(2θ) = 0. Let ￿ > 0. We want to show that there exists a δ such that when |θ − π | < δ we 4 must have | cos(2θ) − cos( π )| = | cos(2θ)| < ￿. This δ will almost always depends on 2 the size of ￿. It is our job to figure out what the correct sort of relationship is. What i write next is not the correct way to write the proof. It is an explanation of how to figure out the correct relationship that you will then use in the proof. So, if we want | cos(2θ)| < ￿, we can deduce: −￿ < cos(2θ) < ￿ 1 1 arccos(−￿) < θ < arccos(￿) 2 2 1 π 1 (π − arccos(￿)) − < θ − π < arccos(￿) − 4 2 4 2 π1 1 − arccos(￿)) < θ − π < arccos(￿) − 4 42 2 π 1 So, if δ = | 4 − 2 arccos(−￿)| we must have that cos(2θ) < ￿. Now the correct way to write this proof is as follows: Let ￿ > 0. If we require that π π1 |θ − | < | − arccos(−￿)| 4 42 then we get that: π1 1 − arccos(￿)) < θ − π < arccos(￿) − 4 42 2 1 π 1 (π − arccos(￿)) − < θ − π < arccos(￿) − 4 2 4 2 1 1 arccos(−￿) < θ < arccos(￿) 2 2 −￿ < cos(2θ) < ￿ π 4 π 4 π 4 π 4 So, we have shown that for any ￿ > 0, there exists a δ such that when |θ − π | < δ , 4 cos(2θ) < ￿. ￿ ...
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This note was uploaded on 02/10/2011 for the course APPM 2350 taught by Professor Adamnorris during the Fall '07 term at Colorado.

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