Math 127(3) - Warren D. Smith - Homework #3 - due Tues 11 Feb 2003 ------------------------------------------------------------------ ANSWERS Finish reading chapters 12 and 13. Don't need to read the 13 stuff on planetary orbits, at least not yet. Start 14. 13.1 ("vector functions") 8: find t-derivative of this vector-valued fn of t: ( (t+1)/(t-1), t exp(2t), sec(t) ) ANS We differentiate the 3 components individually the first by the quotient rule followed by simplification, the second by product rule, the third by formula (or quotient rule). Result: 2 ( -2/(t-1) , exp(2 t) + 2 t exp(2 t) , sec(t) tan(t) ) That first component again in slow-mo: To differentiate (x+1)/(x-1) with respect to x, use quotient rule 2 [T(x)/B(x)]' = [T'(x) B(x) - T(x) B'(x)]/B(x) getting with T(x)=(x+1) and B(x)=(x-1), T'(x)=1, B'(x)=1, so result is 2 2 [(x+1)/(x-1)]' = [x-1 - (x+1)]/(x-1) = -2/(x-1). That third component again in slow-mo: 2 sec(x) = 1/cos(x), derivative is sin(x)/cos(x) = sec(x) tan(x). by quotient rule (or more simply power = -1 rule with chain rule, note get two - signs which yield a + sign inside here...). 12: find second t-derivative of 1/2 3/2 ( t , t , ln(t) ) ANS We differentiate the 3 components individually once: -1/2 1/2 -1 ( (1/2) t , (3/2) t , t ) then again: -3/2 -1/2 -2 ( (-1/2)(1/2) t , (1/2) (3/2) t , -t ) which is -3/2 -1/2 -2 ( (-1/4) t , (3/4) t , -t ). 14. integrate from t=0 to pi, dt: (sin t, cos t, t) ANS We integrate each component individually to get 2 | pi ( -cos(t), sin(t), t / 2 ) | | 0 which is 2 ( 2, 0, pi / 2 ) 40: find a parameterized curve (vector-valued fn of t) describing the curve 2 2 (x-1) + y = 1 and going (a) clockwise (b) anticlockwise. ANS x = 1 + cos(t), y = sin(t) goes anticlockwise x = 1 + cos(-t), y = sin(-t) goes clockwise 2 2 Pythagoras sin(t) + cos(t) = 1 makes this work. Other answers are possible but these are the simplest. 13.2 ("differentiation formulas") 4: find the 1st and 2nd derivative with respect to t of this vector-valued fn (. means vector dot product) 2 2 [ (t , -1, 0) . (1, -t , 0) ] (1,0,0) Evaluating the dot product, 2 2 this is [2 t ] (1,0,0) = (2 t , 0, 0). So the first deriv is (4 t, 0, 0) This can also be done using the dot product "product rule" (A.B)' = A'.B + A.B' but it seems more painful to do it that way. And the second deriv is (4 , 0, 0). 8: and this one (X means vector cross product): 2 3 (t, -t , 1) X (1, t , 5t) Evaluating the cross product we have 3 2 4 2 ( - 6 t , 1 - 5 t , t + t ) The first derivative is 2 3 ( -18 t , - 10 t , 4 t + 2 t ) This can also be done using the cross product "product rule" (A X B)' = A' X B + A X B' but it seems more painful to do it that way since you have to do more cross products. (It is nicer to do it that way if staying at the A,B vector level of abstraction, but not at the present nitty-gritty full-details-of-all-coordinates level.) The second derivative is 2 ( -36 t , - 10 , 12 t + 2 ) WDS3. Consider the plane curve whose curvature (=1/RadiusOfCurvature) c is, as a function of arc length s along the curve, c(s)=s. Assume this curve starts out at the origin x=y=0 moving in the x-axis direction. Write this curve in parametric form (that is, x,y are functions, which you give in closed form, of some other variable t; also you can, if desired, just write y as a function of x). ANS: This curve as you go further along it (s increases) gets more and more spirally since larger and larger curvature. We have angle to the x-axis = theta = integral from 0 to s of c(s) ds = integral from 0 to s of s ds = s^2 / 2. (since c(s) = d(theta)/d(s) by definition of curvature). We now write (x,y) = integral from 0 to s ( cos(theta(s)), sin(theta(s) ) ds 2 2 = integral from 0 to s ( cos(s / 2), sin(s /2 ) ds You should have gotten at least this far. Now: Can these integrals be done? Well... unfortunately, no, not in terms of the elementary functions calc I and calc II students have usually learned. However, they CAN be expressed in terms of non-elementary functions called "Fresnel integrals" or "Fresnel functions" (same Fresnel who invented "Fresnel lens") as you'd see if you looked up the integral of 2 2 cos(x ) dx and sin(x ) dx in a good table of integrals [Gradshteyn and Ryzhik: Table of Integrals, Series, and Products section 8.25 in my edition] or looked up Fresnel integrals in Abramowitz&Stegun: "Handbook of mathematical functions" [probably the world's most useful math book, see chapter 7 for Fresnel integrals; this book available ONLINE from my math127 web page]. Definitions: 2 FresnelC(x) = integral from t=0 to t=x of cos( pi t / 2 ) dt 2 FresnelS(x) = integral from t=0 to t=x of sin( pi t / 2 ) dt (You have to change variables from s to Rs, where R=SquareRootOf(pi), to use these formulas from the book. The need to change variables by such rescalings, etc, is common when tables of integrals are used because the book's formulas are usually not scaled the same as yours.) So the result of doing our integral with the aid of these book formulas is (x,y) = ( R FresnelC( s/R ) , R FresnelS( s/R ) ) where I write R=SquareRootOf(pi) for short. This curve is called a "Fresnel spiral." WDS4. A thrown object describes a parabolic curve of the form 2 x = A t y = B t - C t where t is time since throw it, and A,B,C are positive constants. Find the arc length of this curve, i.e. the total distance the object travels, as a function of t. (Should be 0 when t=0 and should be positive when t>0.) If you cannot do the integral leave it unevaluated and get less credit, but you should be able to do the integral with some work to get a closed form in terms of A,B,C, and t. ANS: The infinitesimal element of arc length is 2 2 2 2 ds = SquareRootOf( dx + dy ) = SquareRootOf( A + (B - 2Ct) ) dt. To find the arc length s we integrate form 0 to t: 2 2 s = integral from 0 to t of SquareRootOf( A + (B - 2Ct) ) dt 2 2 2 2 = integral from 0 to t of SquareRootOf( A + B - 4CBt + 4C t ) dt. You should have gotten at least this far. Now the question is: how can we do this integral? Well, again, by far the easiest way is to look it up in a table of integrals. There is no reason to be shy about trying to take advantage of years of mathematical labor in these books. Even a fairly weak table ought to have this one, but it will be written in the form [Grad&Rhyz section 2.25 for example, formula 2.262 my edition] 2 integral of SquareRootOf( a x + b x + c ) dx 2 2 2 and you will of course have to realize that a=4C , b=-4CB, c=A + B (for us) to use it. If you do not have a table of integrals, then you can still do these kind of integrals by various changes of variables, for example "Euler's change of variables" converts the integral to a purely rational function which can be handled by partial fractions... and certain trig and hyperbolic-trig substitutions will also work... Basically this kind of integral is almost the same as the kind of integral 2 integral ( 1 - x ) dx you get when you are finding the area of a semicircle (or a sub-slab thereof) except for a linear change of variables and a change of the sign of x-squared, which causes certain trig functions to need to be replaced by analogous hyperbolic functions. So when you think of it that way there is no reason at all to be scared of it and you will know in your bones it can be done. For simplicity in the equations that follow, define 2 2 R(t) = SquareRootOf( A + (B - 2Ct) ), 2 2 Q(t) = -4C (B-2Ct), P = 16C = (4C) and 2 2 D = 16 C A , so that s = integral from 0 to t of R(t) dt. We have from tables (and the point of making these abbreviations R,Q,P,D was so I could now just copy the answer easily from the tables... your tables of integrals may differ in which case the easiest abbreviations for you to use may differ slightly) 2 integral of R(t) dt = Q(t) R(t)/P + A /(4C) Arcsinh( Q(t)/SquareRoot(D) ) (Incidentally: as a sanity check, I differentiated the right hand side and got, after some simplifications, R(t).) so that at last 2 s = Q(t) R(t)/P + A /(4C) Arcsinh( Q(t)/SquareRoot(D) ) 2 - Q(0) R(0)/P - A /(4C) Arcsinh( Q(0)/SquareRoot(D) ). Incidentally: Do not be afraid to use answers which are several equations (a bunch of abbreviations leading up to the main result). This kind of answer is often easier to deal with than what you get by fully expanding everything out into 1 giant formula. If you did that in this case, the result would be large and messy and the nice visible structure of it would be lost from view. The goal is to get a formula that can be evaluated with a computer or calculator, and this kind of formula satisfies that goal fine.) ----------------------------------------------------------------