course Mth 275
Section 11.4:This is getting more difficult. The material in the section itself is clear and understandable. However a couple of the problems gave me difficulty. 51(b) says to use the computer to compute the length of a curve in R3. I was unable to determine how to do this with Derive 6.
t goes from 0 to 2 pi. The arc distance corresponding to an interval `dt will be sqrt( x'^2 + y'^2 + z'^2) `dt. You can easily calculate the derivatives with or without DERIVE, and write down this expression. Just integrate this expression with respect to t, from t = 0 to 2 pi.
The curve starts at (1, 0, 0) and ends up at about (.1, 0, 1), sort of spiraling upward as it moves toward and around the z axis. Total length can be very roughly estimated in comparison to the circumference of a circle of radius 1, which would be about 6. This curve spirals in much closer to the origin and doesn't rise much; I'm guessing that the length would be between 1 and 2.
I just realized, when I started answering your next question, that I have DERIVE on my computer. Keying in the functions, letting DERIVE calculate the derivatives because it's late and I'd probably mess it up, integrating the square root of the sum of the squares of these derivatives between 0 and 2 pi yields 1.7997 and change. The integral was
int( ((- ê^(- 3·t)·(3·COS(t) + SIN(t)))^2+(ê^(- 3·t)·(COS(t) - 3·SIN(t)))^2+(13·(40 - t^2)/(t^2 + 40)^2)^2)^.5,t,0,2 pi)
See if you agree.
Number 57 asks for a procedure to find an equation for the osculating circle to graph C of R(t) at point P(a, b). I would find T(t), N(t), and k, then use 1/k for r, and find the center of the circle by going r distance along N(t) from P. Part B wants this done for point P(32,122) on C defined by x = 32t and y = 16t^2 ?4. I computed t = 1, but I got hung up in the calculations for N(t). Also I? not sure how to find the center from r and N(t).
If the curve is in the plane it's a lot easier than in 3 dimensions.
N(t) is always one of the fun calculations in multivariable calculus. x ' = 32, y ' = 32 t, so v = 32 i + 32 t j and T(t) = v / || v || = (32 i + 32 t j) / (32 sqrt(1 + t^2) ) = (i + t j) / sqrt(1 + t^2). The unit normal is perpendicular to this so is either (t i - j) / sqrt(1+t^2) or (-t i + j ) / sqrt ( 1 + t^2). From the upward curvature of the path it will be the latter.
But that's cheating. By the quotient rule T ' (t) = (j sqrt(1 + t^2) - t / sqrt(1 + t^2) ( i + t j) ) / (1 + t^2) = (j - i t) / (1 + t^2)^(3/2). The magnitude of this vector is (1 + t^2) / ( 1 + t^2)^(3/2) = 1 / (1 + t^2), to T ' (t) / || T ' (t) || = (j - i t) / (1 + t^2)^(3/2) / ( 1 / (1 + t^2) ) = (j - i t) / sqrt(1 + t^2), as claimed above.
To get the center from the origin, just go to the point P; the vector from origin to P is (x(t), y(t)) = (32 t, 16 t^2 - 4). Then follow the direction of the unit normal for distance r, which you accomplish by adding r * N(t) to the position vector. I think from what you say that you can find the expression for r, but if not let me know.
Number 60 asks for proof of the formula for torsion. I got nowhere with this one. The second part of the problem asks for the torsion for the helix R(t) = (a cos t)i + (a sin t)j + (bt)k. I came up with (a^2b)/ sqrt (a^2*b^2 + a^4). Is this correct? Finally, problems 58 and 59 state that torsion is a measure of the amount of twisting at each point on the trajectory. Is this a measure of the speed (magnitude) of the change in direction of the moving body at a given point?
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I'll have to defer my response on this last question, which fits in nicely with some of my previous responses. Remind me if I don't post anything by tomorrow night.