jf 00 (x)j ds (x) = [1 + (f 0 (x)) 2 ] 3=2 (t) = jjr0 (t) r 00 (t)jj jjr 0 (t)jj 3

Transcription

1 M73Q Multivariable Calculus Fall 7 Review Problems for Exam The formulas in the box will be rovided on the exam. (s) dt jf (x)j ds (x) [ + (f (x)) ] 3 (t) jjt (t)jj jjr (t)jj (t) jjr (t) r (t)jj jjr (t)jj 3. True or False? Circle ONE answer for each. Hint: For eective study, exlain why if `true' and give a counterexamle if `false.' (a) T or F: If a?b and b?c; then a?c: Solution: False. Let a i; b j; and c < ; ; > : Then a b i j and b c < ; ; > < ; ; > : However, a c < ; ; > < ; ; > : (b) T or F: If a b ; then jja bjj jjajjjjbjj: Solution: True. Assume both a and b are nonero. (If either is ero the result is obvious). If is the angle between a and b; then since a and b are orthogonal. jja bjj jjajjjjbjj sin jjajjjjbjj sin jjajjjjbjj: (c) T or F: For any vectors u; v in R 3 ; jju vjj jjv ujj: Solution: True. If is the angle between u and v; then jju vjj jjujjjjvjj sin jjvjjjjujj sin jjv ujj: (Or, jju vjj jj v ujj j jjjv ujj jjv ujj:) (d) T or F: The vector < 3; ; > is arallel to the lane 6x y + 4 : Solution: False. A normal vector to the lane is n < 6; ; 4 > : Because < 3; ; > n; the vector is arallel to n and hence erendicular to the lane. (e) T or F: If u v ; then u or v : Solution: False. For examle i j but i 6 and j 6 : (f) T or F: If u v ; then u or v : Solution: False. For examle i i but i 6 : (g) T or F: If u v and u v ; then u or v : Solution: True. If both u and v are nonero, then u v imlies u and v are orthogonal. But u v imlies that u and v are arallel. Two nonero vectors can't be both arallel and orthogonal, so at least one of them must be : (h) T or F: The curve r(t) ; t ; 4t is a arabola. Solution: True. Parametric equations for the curve are x ; y t ; 4t; and since t 4 we have y t or y ; x : This is an equation of a arabola in the y lane. 4 6

2 M73Q Multivariable Calculus Review Problems for Exam - Page of 9 (i) T or F: If (t) for all t; the curve is a straight line. Solution: True. Notice that (t) () jjt (t)jj () T (t) for all t: But then T(t) C; a constant vector, which is true only for a straight line. (j) T or F: Dierent arameteriations of the same curve result in identical tangent vectors at a given oint on the curve. Solution: False. For examle, r (t) < t; t > and r (t) < t; t > both reresent the same lane curve (the line y x), but the tangent vector r (t) < ; > for all t; while r (t) < ; > : In fact, dierent arametriations give arallel tangent vectors at a oint, but their magnitudes may dier.. Which of the following are vectors? (a) Vector Scalar Nonsense [(a b)c] a (b) Vector Scalar Nonsense c [(a b) c] (c) Vector Scalar Nonsense c [(a b)c] (d) Vector Scalar Nonsense (a b) c 3. Which of the following are meaningful? (a) Meaningful Nonsense jjwjj(u v) (b) Meaningful Nonsense (u v) w (c) Meaningful Nonsense u (v w) 4. Find the values of x such that the vectors < 3; ; x > and < x; 4; x > are orthogonal. Solution: For the two vectors to be orthogonal, we need < 3; ; x > < x; 4; x > : That is, 3(x) + (4) + x(x) ; or x or x 4: 5. Find the decomosition a a kb + a?b of a < ; ; > along b < ; ; 3 > : Solution: a kb a b b b b ; 7 ; : ; 7 ; : Finally, a?b a a 7 kb < ; ; > 3 7 ; 7 ; 7 6. Let a ; ; b ; ; and u < 3; > : (a) Show that a and b are orthogonal unit vectors. Solution: a b and jjajj + ; and jjbjj +( ) : (b) Find the decomosition of u along a: Solution: u jja (u a)a 3 ; 3 ; 3 : 3 3 ; : Thus, u?a < 3; > 3 ; 3 (c) Find the decomosition of u along b:

3 M73Q Multivariable Calculus Review Problems for Exam - Page 3 of 9 Solution: Similarly, u jjb (u b)b < 3; > ; Thus, u?b < 3; 3 > ; 3 3 ; 3 u?a : ; 3 3 ; : 7. (a) Find an equation of the shere that asses through the oint (6, -, 3) and has center (-,,). Solution: Use the distance formula to nd the distance between (6, -, 3) and (-,,). Then, the equation for the circle is (x + ) + (y ) + ( ) 69: (b) Find the curve in which this shere intersects the y-lane. Solution: The intersection of this shere with the y lane is the set of oints on the shere whose x coordinate is. Putting x in to the equation, (y ) + ( ) 68; which reresents a circle in the y lance with center (; ; ) and radius 68: 8. For each of the following quantities (cos ; sin ; x; y; ; and w) in the icture below, ll in the blank with the number of the exression, taken from the list to the right, to which it is equal. a y w θ x b cos 5. sin 4. x 3. y 3 4. a b jjajj a b jjbjj jja bjj jjbjj jja bjj jjajjjjbjj 7 5. ja bj jjajjjjbjj w 6 6. jjb ajj 7. (b a) b jjbjj 9. Find an equation for the line through (4; ; ) and (; ; 5): Solution: The line has direction v < 3; ; 3 > : Letting P (4; ; ); arametric equations are x 4 3t; y + t; + 3t:

4 M73Q Multivariable Calculus Review Problems for Exam - Page 4 of 9. Find an equation for the line through ( ; ; 4) and erendicular to the lane x y + 5 : Solution: A direction vector for the line is a normal vector for the lane, n < ; ; 5 >; and arametric equations for the line are x + t; y t; 4 + 5t:. Find an equation of the lane through (; ; ) and arallel to x + 4y 3 : Solution: Since the two lanes are arallel, they will have the same normal vectors. Then we can take n < ; 4; 3 > and an equation of the lane is (x ) + 4(y ) 3( ) :. Find an equation of the lane that asses through the oint ( ; 3; ) and contains the line x(t) t; y(t) 4t; (t) + t: Solution: To nd an equation of the lane we must rst nd two nonarallel vectors in the lane, then their cross roduct will be a normal vector to the lane. But since the given line lies in the lane, its direction vector a < ; 4; > is one vector arallel to the lane. To nd another vector b which lies in the lane ick any oint on the line [say ( ; ; ); found by setting t ] and let b be the vector connecting this oint to the given oint in the lane ( ; 3; ): (But beware; we should rst check that the given oint is not on the given line. If it were on the line, the lane wouldnt be uniquely determined. What would n then be when we set n a b?) Thus b < ; 3; > and n a b < + 3; ; 6 >< 3; ; 6 > and an equation of the lane is 3(x + ) + 6( ) or x + 3: 3. Find the oint at which the line x(t) t; y t; (t) + t and the lane x + y intersect. Solution: Substitute the line into the equation of the lane. x+y ) +t ( t)+t ) t ) the oint of intersection is (; ; ): 4. (a) Find an equation of the lane that asses through the oints A(; ; ); B( ; ; ); and C(; 3; 4:)! Solution: The vector AB < 3; ; 9 > and! AC < ; ; 5 > lie in the lane, so n! AB! AC < 8; 4; 8 > or equivalently, < ; 3; > is a normal vector to the lane. The oint A(; ; ) lies on the lane so an equation for the lane is (x ) + 3(y ) + ( ) : (b) A second lane asses through (; ; 4) and has normal vector < ; 4; 3 > : Find an equation for the line of intersection of the two lanes. Solution: The oint (,,4) lies on the second lane, but the oint also satises the equation of the rst lane, so the oint lies on the line of intersection of the lanes. A vector v in the direction of this intersecting line is erendicular to the normal vectors of both lanes, so take v < ; 3; > < ; 4; 3 >< 5; 5; > or just use < ; ; > : Parametric equations for the line are x + t; y t; 4 + t: 5. Provide a clear sketch of the following traces for the quadratic surface y x + + in the given lanes. Label your work aroriately. x ; x ; y ; y ; :

5 M73Q Multivariable Calculus Review Problems for Exam - Page 5 of 9 y x -axis x y -axis -axis -axis Solution: 6. Match the equations with their grahs. Give reasons for your choices. (a) II 8x + y + 3 I sin x + cos y (b) (c) IV sin (d) III ey + x + y 7. Find a vector function that reresents the curve of intersection of the cylinder x + y 6 and the lane x + 5: Solution: The rojection of the curve C of intersection onto the xy lane is the circle x + y 6; : So we can write x 4 cos t; y 4 sin t; t : From the equation of the lane, we have 5 x 5 4 cos t; so arametric equations for C are x 4 cos t; y 4 sin t; 5 4 cos t; t ; and the corresonding vector function is r(t) < 4 cos t; 4 sin t; 5 4 cos t >; t : 8. Find an equation for the tangent line to the curve x sin t; y sin t; and sin 3t at the oint (; 3; ):

6 M73Q Multivariable Calculus Review Problems for Exam - Page 6 of 9 Solution: The curve is given by r(t) < sin t; sin t; sin 3t >; so r (t) < cos t; 4 cos t; 6 cos 3t > : The oint (; 3; ) corresonds to t 6; so the tangent vector there is r (6) < 3; ; > : Then the tangent line has direction vector < 3; ; > and includes the oint (; 3; ); so arametric equations are x + 3t; y 3 + t; : 9. A helix circles the axis, going from (; ; ) to (; ; 6) in one turn. (a) Parameterie this helix. Solution: r(t) < cos t; sin t; 3t > : (Note that revolution is ; so (3) 6:) (b) Calculate the length of a single turn. Solution: For t ; jjr (t)jj : Thus s (c) Find the curvature of this helix. Z 3dt 3(): Solution: The unit tangent vector is T(t) 3 < sin t; cos t; 3 >; so T (t) 3 < cos t; sin t; >. Thus, jjt (t)jj 3 : Since jjr (t)jj 3; 3. (a) Sketch the curve with vector function r(t) ht; cos t; sin ti ; t : Solution: The corresonding arametric equations for the curve are x t; y cos t; sin t: Since y + ; the curve is contained in a circular cylinder with axis the x axis. Since x t; the curve is a helix. (b) Find r (t) and r (t): Solution: Since r(t) ht; cos t; sin ti ; r (t) h; sin t; cos ti ; and r (t) ; i cos t; sin t :. Which curve below is traced out by r(t) sin t; cos 4 t; t ; t : Solution: Grah. Note that r() < ; ; > :. Find a oint on the curve r(t) t + ; t ; 5 where the tangent line is arallel to the lane x + y 4 5:

7 M73Q Multivariable Calculus Review Problems for Exam - Page 7 of 9 Solution: The lane has normal vector n < ; ; 4 > : Since r (t) < ; 4t; >; we want < ; 4t; > < ; ; 4 > : That is, + 8t ; and r( ) 8 < 7 8 ; 63 4 ; 5 > : 3. Let r(t) t; (e t )t; ln(t + ) : (a) Find the domain of r: Solution: The exressions t; (e t )t; and ln(t + ) are all dened when t : Thus, t ; t 6 ; and t + > ) t > : Finally, the domain of r is ( ; ) [ (; ]: (b) Find lim r(t): t! Solution: lim r(t) t! D ; ; E : Note that in the y comonent we use l'hosital's Rule. (c) Find r (t): D Solution: r (t) E ; te t e t + t t ; : t+ 4. Suose that an object has velocity v(t) 3 + t; sin(t); 6e 3t at time t; and osition r(t) < ; ; > at time t : Find the osition, r(t); of the object at time t: Z Solution: r(t) v(t)dt Z 3 + tdt; Z sin(t)dt; Z 6e 3t dt D( + t) 3 + c ; cos(t) + c ; e 3t + c 3 E : Thus, r() < + c ; + c ; + c 3 >< ; ; >) c ; c ; c 3 : So, r(t) D( + t) 3 ; cos(t) + ; e 3t E : 5. If r(t) t ; t cos t; sin t ; evaluate Z r(t)dt: Solution: Z r(t)dt Z Z Z t dt; t cos tdt; sin tdt 3 ; ; : 6. Find the length of the curve: x cos(t); y t 3 ; and sin(t); t : Solution: s 7 : Z s dx + dt dy dt + d dt dt Z q 6 sin (t) + 6 cos (t) + 9t dt Z 6 + 9tdt 7. Rearameterie the curve r(t) < e t ; e t sin t; e t cos t > with resect to arc length measured from the oint (; ; ) in the direction of increasing t: Solution: The arametric value corresonding to the oint (; ; ) is t : Since r (t) < e t ; e t (cos t+ sin t); e t (cos t sin t) >; jjr (t)jj e t + (cos t + sin t) + (cos t sin t) 3e t ; and s(t) R t eu 3du D E 3(e t ) ) t ln + s : 3 Therefore, r(t(s)) + s; 3 + s 3 sin ln + s ; 3 + s 3 cos ln + s : 3

8 M73Q Multivariable Calculus Review Problems for Exam - Page 8 of 9 8. Find the tangent line to the curve of intersection of the cylinder x + y 5 and the lane x at the oint (3; 4; 3): Solution: Let x(t) 5 cos t; y(t) 5 sin t; and (t) 5 cos t; so that r(t) < 5 cos t; 5 sin t; 5 cos t > and r (t) < 5 sin t; 5 cos t; 5 sin t > : When (x; y; ) (3; 4; 3); x(t) (t) 5 cos t 3; and y(t) 5 sin t 4; so r (t ) < 4; 3; 4 > is a direction vector for the line. The tangent line has arametric equations x(t) 3 4t; y(t) 4 + 3t; and (t) 3 4t: ANOTHER SOLUTION: Let x(t) t; y 5 x 5 t ; and (t) x(t) t: Then r(t) D E case, r t (t) ; 5 t ; : When x 3; t 3; and r (3) < ; 3 arametric equations x(t) 3 + t; y(t) 4 t; and (t) 3 + t: 4 D t; (5 t) ; t E : In this 3 ; > : So the tangent line has 4 9. For the curve given by r(t) 3 t3 ; t ; t ; nd (a) the unit tangent vector ; t; > : Solution: T(t) r (t) jjr (t)jj < t ; t; > t 4 + t + < t (t + ) (b) the unit normal vector Solution: T (t) < 4t; 4 t ; 4t > (t + ) ) jjt (t)jj 4(t4 + 4t + 4) (t + ) t + < t; t ; t > and N(t) t + (c) the curvature Solution: (t) jjt (t)jj jjr (t)jj (t + ) 3. A article moves with osition function r(t) < t ln t; t; e t >. Find the velocity, seed, and acceleration of the article. Solution: v(t) r (t) < + ln t; ; e t > : (t) jjv(t)jj ( + ln t) + + ( e t ) + ln t + (ln t) + e t : a(t) v (t) < t ; ; e t > : 3. A article starts at the origin with initial velocity < ; ; 3 > and its acceleration is a(t) < 6t; t ; 6t > : Find its osition function. Z Z Z Z Solution: v(t) a(t)dt 6t dt; t dt; 6t dt 3t ; 4t 3 ; 3t +C; but < ; ; 3 > v()+c; so C < ; ; 3 > and v(t) 3t + ; 4t 3 ; 3t + 3 Z : r(t) v(t)dt t 3 + t; t 4 t; 3t t 3 D: But r() ; so D ; and r(t) t 3 + t; t 4 t; 3t t 3 : 3. A ying squirrel has osition r(t) t + t ; t; + t at time t: Comute the following at time t : (a) The velocity at time t ; v() < ; ; >.

9 M73Q Multivariable Calculus Review Problems for Exam - Page 9 of 9 Solution: v(t) r (t) < + t; ; t > : (b) The seed at time t ; () 3. Solution: jjv(t)jj: jjv()jj : 33. Consider the vector valued function r(t) describing the curve shown below. Put the curvature of r at A; B and C in order from smallest to largest. Draw the osculating circles at those oints. Solution: B, A, C.