Homwework JWR. Jan Problem numbers refer to the do Carmo text.

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1 Homwework JWR Jan Problem numbers refer to the do Carmo text The curve αs) cos s), sin s)) coss), sins)) parameterizes the circle x +y 1 in the clockwise orientation.. 1.-Thedistanceformthepointαt) R n totheoriginisft) α t). At a point where this distance assumes its minimum, the derivative of the function must vanish. But f t) αt) α t) αt) and this vanishes when the vectors αt) 0 and α t) are orthogonal Let αt) xt),yt),zt)). Then α t) x t),y t),z t)). If α t) 0 for all t, then α t) is a constant vector, say α t) v 0 and hence integrating once more) αt) tv 0 +p 0 for some p 0 x 0,y 0,z 0 ). This argument works in n dimensions.) The curve αt) 3t,3t,t 3 ) has velocity vector vt) 3,6t,6t ). The vector u 1,0,1) point along the line y 0, x z. The cosine of the angle θ between these two vectors is defined by cosθ 3+6t 9+36t +36t 4 cosπ/4) where we used the formulas 3+6t ) 9+36t +36t 4 and cosπ/4) 1/. Note: The wording of the problem seems to suggest that all the tangent lines to the curve α intersect the line y 0, x z but this is not correct. The tangent line to the curve α at the point αt) has the parametric equations x,y,z) αt)+rα t) 3t+3r,3t +6rt,t 3 +6rt ). Here the variable r parameterizes the line.) The tangent line intersects the plane y 0 when r t/, but 3t + 3r t 3 + 6rt for this value of r. Evidently, when do Carmo talks about the angle between two lines he means the angle between two vectors along the lines. 1

2 Let α : 1, ) R be defined by αt) xt),yt) ) 3t 3t ) 1+t 3, 1+t 3 gt),gt)t ), gt) : 3t 1+t 3. The derivative α t) is given by α t) g t),g t)t+gt) ), g t) 31+t3 ) 9t 3 1+t 3 ) 31 t3 ) 1+t 3 ). Now g0) 0 and g 0) 3 so α 0) 3,0) which shows that the trace C of α is tangent to the x-axis at the point α0). Now it is easy to see that lim αt) lim t t α t) 0,0) because in every fraction that appears, the degree of the numerator is smaller than the degree of the denominator. Before discussing part c) I ll make a preliminary remark. Every line in the plane has an equation of form a 0 x+b 0 y +c 0 0 where either a 0 0 or b 0 0 or both). Multiplying the vector a 0,b 0,c 0 ) by a nonzero constant does not change the line, so the line does not determine the vector a 0,b 0,c 0 ) uniquely. The vector b 0,a 0 ) is perpendicular to the line at each point so there are only two equations for the line with a unit normal vector, namely ±a 0 x + b 0 y + c 0 ) 0 where a 0 + b 0 1. The definition for what it means for a parameterized family of lines to approach a line Lt) {x,y) : at)x+bt)y +ct) 0} L 0 {x,y,) : a 0 x+b 0 y +c 0 0} as t t 0 is that there is a choice of signs µt) ±1 such that lim µt) at),bt),ct)) t t 0 at) +bt) a 0,b 0,c 0 ). This implies a 0 + b 0 1.) If we want to use language carefully, we should make this explicit before saying something like do Carmo says in part c) of the problem. Now to do part c) we should compute the limit as t 1 of the unit normal vector nt) at),bt)) : y t), x t)) y t) +x t)

3 to the curve αt) xt),yt)). Now x t) g t) and y t) g t)t +gt) so an explicit formula for nt) is gt)+g t)t, g t) ) nt) gt)+g t)t) +g t) ht)+t, 1) ht)+t) +1, gt) ht) : g t). Here we used that g t) > 0 for t 1,0).) Now ht) gt) g t) )/ 3t 31 t 3 ) ) 1+t 3 1+t 3 ) ) 3t 1+t 3 ) ) 1+t 3 31 t 3 ) t1+t3 ) 1 t 3 so lim t 1 ht) 0 and hence lim t 1 nt) 1, 1 ). The tangent line to C at αt) has equation nt) p αt)) 0 where p x,y). This expands to at)x+bt)y +ct) 0 where at) ht)+t ht)+t) +1, bt) t ht)+t) +1, and 3tht)+t) 3t ct) at)xt) bt)yt) ht)+t) +1 1+t 3 ) 3t ht)+t) +11 t 3 ) so the limiting tangent line as t 1 has equation 1 x 1 y 1 0. This simplifies to x + y as in Figure 1-10 in do Carmo. The dashed curve in the figure is defined by the same formula with t, 1) The curve αt) e t cost,e t sint) has velocity vector and speed α t) e t. α t) e t cost sint, sint+cost) A equation for the line through the three points p i x i,y i,z i ) is p3 p 1 ) p 3 p ) ) p p 3 ) 0, p x,y,z). This equation is linear, i.e. it can be written as Ax+By +Cz +D 0 where A y 3 z 1 z 1 y 3 ), B y 3 z z 3 y ), C x 3 y y 3 x ), and D det x 1 x x 3 y 1 y y 3. z 1 z z 3 3

4 The three points p i obviously lie on the surface p p1 ) p p ) ) p p 3 ) 0, but it appears that this surface is cubic, not linear. However the surface is a plane. To see this write the triple product as a determinant p p1 ) p p ) ) p p ) 3 det x x 1 x x x x 3 y y 1 y y y y 3. z z 1 z z z z 3 Subtracting the third column from the first and second gives p p1 ) p p ) ) p p ) x 3 x 1 x 3 x x x 3 3 det y 3 y 1 y 3 y y y 3 z 3 z 1 z 3 z z z 3. This is the same as the linear equation above Assume that two maps u,v : I R 3 satisfy a differential equation u au+bv, v cu av where a,b,c are constants. Then u v) u v+u v au+bv) v+u cu av) 0 as v v u u 0 and the wedge product is distributive The helix αs) acosθ,asinθ,bθ), θ s c, a +b c has derivative α s) c 1 asinθ,acosθ,b) with length α s) a +b / c so α is parameterized by arc length. Since α s) ac cosθ,sinθ,0) the curvature is κ ac, and the Frenet trihedron is t c 1 asinθ, acosθ, b), n cos θ, sin θ, 0), b c 1 bsinθ, bcosθ, a). Since b bc cosθ,sinθ,0) the torsion is τ bc by the third Frenet formula. A point p lies on the osculating plane at αs) if and only if. When p x,y,z) this expands to b p αs)) 0. bsinθ)x asinθ)+bcosθ)y acosθ) az bθ) 0 4

5 and simplifies to bsinθ)x+bcosθ)y az +ab1+θ) 0. The line through αs) along the normal ns) has parametric equation γr) αs)+rns) a+r)cosθ,a+r)asinθ,bθ ) andmeetsthe z-axiswhen r a. Thecosineofthe anglebetween unit tangent t and the vector 0,0,1) is the constant t 0,0,1) c 1 asinθ,acosθ,b) 0,0,1) a/c Suppose that the curve α αs) is parameterized by arc length and all its normal lines pass through the point o R 3. Then there is a function r rs) such that αs)+rs)ns) o. Differentiating and applying the Frenet formulas we get t+r n+r κt+τb) 0. But the vectors t,n,b are linearly independent so r 0, 1 rκ, and τ 0. Hence r is constant, r 1/κ, τ 0, b is constant, and the curve lies in the plane through o perpendicular to b. The penultimate assertion uses the third Frenet equation b τn. The last statement follows from the fact that the osculating plane has equation b p o) 0 and the fact that b is constant.) 5

6 The following problem is important but was not assigned) Let the position of a particle at time t be given by αt) βσt)) where β is parameterized by arclength and σt) αt) is the speed of the particle. Denote differentiation with respect to t by an overdot and differentiation with respect to s by a prime so if gt) fσt)) we have ġt) f σt)) σt) by the chain rule. By the Frenet equations α σt α σt+ σ t σt+ σ κn... α... σt+ σ σt + σ σκ+ σ 3 κ )n+ σ 3 κn... σ σ 3 κτ)t+ σ σκ+ σ σκ+ σ 3 κ )n+ σ 3 κτ b From the first two equations we get α α σ 3 κb and hence α α σ κ. Represent all three equations in the matrix form α α σ 0 0 σ κ 0 t n... α σ 3 κτ b and we get α α)... α σ 6 κ τ. Combining gives κ α α σ 3, τ α α)... α α α. A plane curve is a special case: take αt) xt),yt),0). But in the case of a plane curve we can define the unit normal vector by rotation the unit tangent vector clockwise by a right angle. To maintain the second Frenet equation t κn we must allow κ to take negative values. The above equation for κ becomes κ α α σ 3 ẋÿ ẏẍ ẋ +ẏ ) 3/ In terms of the velocity vector v α and the acceleration vector a α the above equations for α and α take the form b σt, a σt+ σ κn. The second equation resolves the acceleration into its tangential and normal componenets and explains what happens when a car goes around a sharp curve in the road. 6