course Mth 275
Section 11.5:The contents of the section make sense, but as usual, a couple of the problems gave me fits. Problems 9 and 12 ask for T, N, AsubT, AsubN given V and A (9: V= <1,-3>, A=<2,5). I had no trouble finding ||V|| = sqrt 10 and ||A|| = sqrt 29 and finding T = V/||V|| = (1/sqrt 10)<1, -3>, but that’s as far as I can get. I’m sure I’m missing something simple here. Number 12 is similar, but is in R3.
A_parallel = (A dot T) * T is the vector component of A in the direction of the velocity.
Since A_parallel + A_normal = A, you can find A_normal as A - A_parallel = A - (A dot T) * T.
#21 asks for AsubN and AsubT given a parabolic path y^2=4x^2 at the instant speed is ds/dt = 20. I tried substituting t = x^2, so x = sqrt t and y = 2 sqrt t. Thus, R(t) = (sqrt t)i + (2sqrt t)j, V(t) = (1/2)t^(-1/2)i + t^(-1/2)j, and ds/dt = sqrt[5/(4t)]. Substituting ds/dt = 20 gives t = 1/320 which contrasts with the book solution of t= (sqrt 399)/8. I’m also curious as to why this equation is a parabolic path, since taking the sqrt of both sides gives y = 2x.
You are right. That wouldn't be a parabolic path. y^2 = 4 x^2 has two solutions for y in terms of x: y = 2 x or y = -2x. Both are linear.
I suspect a typo in the text. Let's assume that the function is y = 4 x^2.
Then dy/dx = 8 x and ds/dx = sqrt(1 + (dy/dx)^2) = sqrt(1 + 64 x^2). This is equal to 20 when 1 + 64 x^2 = 400; solving for x we obtain the book's answer sqrt(399) / 8.
A vector parallel to the velocity at position (x, 4x^2) would be i + 8 x j; the unit tangent vector would be (i + 8 x j) / sqrt(1 + 64x^2).
A unit normal vector would be (-8x i + j) / sqrt(1 + 64 x^2).
The acceleration vector, assuming t = x, is just 8 j. The component of this vector parallel to T is 8 j dot T = 64 t / sqrt(1 + 64 t^2). Its component in the direction of the normal is 8 / sqrt(1 + 64 x^2).