Math 1120 Calculus II for Engineering, Section 03 { Midterm Examination, Spring, Name: Solution UID: Solution Score: 100

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1 Homework Solution Math 0 Calculus II for Engineering, Section 0 { Miterm Examination, Spring, 008 Name: Solution UID: Solution Score: 00 Section I. Multiple Choice Problems. Compute the area of the triangle with two sies represente by u = i + j k an v = i k. Answer. The area of the parallelogram forme by u an v is obtaine by ku vk. The area A of such a triangle is the half of the area insie the parallelogram an so the esire answer is A = ku vk=. A simple computation shows u v = (i + k) ; ku vk = ki + kk = p ; A = p :. Fin the symmetric equation of the line passing through the points P (; ; ) an Q(5; ; 4). Answer. The vector! P Q = h5 ; ; 4 ( )i = h4; 5; 5i is the vector parallel to the line. Thus, by the enition of the line, one possible symmetric equation is as follows: x 4 = y 5 = z + 5 :. Evaluate the integral Z t + 9; t + 8 t. Answer. Z t + 9; t + 8 t = where c = hc ; c i is a constant vector. = Z t + 9 Z t; (t + 8) t = *ln t + 9t + c ; t *ln t + 9t; t + 8t + + hc ; c i = + 8t + c *ln t + 9t; t + 8t + + c; + 4. Compute f y (x; y) of f(x; y) = e xy + x y. Answer. The Chain Rule implies f y (x; y) = @y! x = xe xy x y y :

2 Section II. Multiple Step Problems. The thrust of an airplane's engine prouces a spee of 500 mph in still air. The win velocity is given by h0; 80i. In what irection shoul the plane hea to y ue east (i.e., in the irection of the unit vector i = h; 0i)? Answer. Let a = hx; yi be the airplane's velocity an w the win's velocity an m the esire moving irection. Then, w = h0; 80i an m = hc; 0i, where c is a positive constant. It implies a + w = m; hx; yi + h0; 80i = hc; 0i ; hx + 0; y + 80i = hc; 0i ; i:e:; x + 0 = c; y + 80 = 0; i:e:; x = c 0; an y = 80: Since the airplane's engine prouces the spee of 500 mph, so kak = 500, i.e., 500 = kak = k hx; yi k = k hc 0; 80i k = q (c 0) + 80 ; i:e:; 500 = (c 0) + 80 ; i:e:; c = 0( p 609): Since c shoul be positive, we have c = 0( + p 609) an so x = c 0 = 0 p 609; an y = 80; i:e:; a = D 0 p 609; 80 E = 0 D p 609; 4 E :. For a = h; i an b = h; 4i, compute the followings. () a b Answer. a b = h; i h; 4i = h() (); () (4)i = h0; 5i : () Comp b a. Answer. Comp b a = a b kbk = h; i h; 4i k h; 4i k = 0 5 = :. Consier the circle r(t) = a cos ti + a sin tj with raius a. () Fin the unit tangent vector. Answer. In the component form, r(t) = a cos ti + a sin tj = a hcos t; sin ti. By the enition, the unit tangent vector T (t) is obtaine as follows: r 0 (t) = a h sin t; cos ti ; kr 0 (t)k = jaj = a

3 T (t) = r0 (t) kr 0 (t)k = a h sin t; cos ti a = h sin t; cos ti : () Show that the curvature is the reciprocal of the raius a. Answer. T 0 (t) = h cos t; sin ti = hcos t; sin ti. = kt 0 (t)k kr 0 (t)k = k hcos t; sin ti k a Hence, the curvature is the reciprocal of the raius a. = a ; i:e:; = a : 4. A cone has the height h an the circular base of raius r. The height an raius changes with time t. At the instance in which r = cm, h = cm, it was foun that r h = 0:0 cm/sec an t t = 0:0 cm/sec. At what rate is the cone's volume changing at that instant? (Hint: One may use the formula for the volume, V = r h.) Answer. Since r an h are functions of t, we have V (t) = r (t)h(t) an so V (t) t = t = V (t) t = () r=; h=; r 0 =0:0; h 0 = 0:0 r (t)h(t) = " # r (t)h(t) t # " " # r(t)h(t) r(t) + r (t) h(t) = r(t) h(t) r(t) + r(t) h(t) t t t t [()(0:0) + ( 0:0)] = 0:04: 5. Let c = b a with kak = = kbk. If an are the angles between a an c an between b an c, respectively, then () n the relation between cos an cos, Answer. Without loss of generality, we assume c 6= 0 so that 6= 0. Then, we have cos = cos = a c kak = a b b c kak kak a b kbk = kbk kbk = a b ; = a b : From those two equations, we euce the ratio, cos cos = a b a b = a b a b = ; i:e:; cos cos = : Thus, we conclue cos = cos.

4 () hence, n the relation between an. Answer. From the Calculus I, We recall the formulas: cos A = cos B if an only if A = B + n; n = 0; ; ; : : : ; cos A = cos B if an only if A = B + (n + ); n = 0; ; ; : : : : (F) Hence, the relation between an is = + (n + ), where n = 0; ; ; : : :. Remark: How o we know the formula (F)? cos(x + y) = cos x cos y sin x sin y: (From Calculus I) cos A = cos(b + (n + )) = cos B cos((n + )) sin B sin((n + )) = (cos B)( ) (sin B)(0) = cos B: That is, cos A = cos B if an only if A = B + (n + ), n = 0; ; ; : : :. Asie: Using the formula from Calculus I, sin(x + y) = sin x cos y + cos x sin y; we can also euce the property, sin A = sin B if an only if A = B + (n + ); n = 0; ; ; : : : : sin A = cos B if an only if A = B + n + ; n = 0; ; ; : : : : sin A = cos B if an only if A = B + n + ; n = 0; ; ; : : : : In summary, for n = 0; ; ; : : :, we have sin A = sin B; cos A = cos B if an only if A = B + n; sin A = cos B if an only if A = B + n + ; sin A = sin B; cos A = cos B if an only if A = B + (n + ); sin A = cos B if an only if A = B + n + : 4

5 Section III. Concept Problems. True or False: In the box, put T for the correct statement an F for the wrong statement. () If kuk an kvk are large, then u v is also large. F Answer. Recall u v = kukkvk cos. Even if kuk an kvk are large, u v can be small when the angle between u an v gives cos <. () For a given vector, there is one unit vector parallel to it. F Answer. For a given vector v, there are two unit vector parallel to it: v kvk an v kvk : \Being parallel" means \pointing the same irection" an \pointing the opposite irection". () The cross prouct of two unit vectors is also a unit vector. F Answer. Recall ka bk = kakkbk sin. Even if kak = an kbk =, the cross prouct a b cannot be the unit vector when sin 6=. For example, a = h; 0; 0i an b = D =; =; = p E are unit vectors. But, So a b is not a unit vector. a b = * 0; + p ; ; ka bk = p : (4) The minimum number of points require to etermine an equation of a plane in the {imensional space is two points. Answer. Three points etermines a triangle. One may say that a triangle is the simplest plane. That is, the minimum number of points require to form a plane is three. F (5) If r(t) is a ierential vector function, then t kr(t)k = r(t) t. F Answer. By the Theorem. (v) on page 805, we recall kr(t)k = (r(t) r(t)) = : We ierentiate both sies. The Chain Rule an the Theorem. (v) on page 805 imply t kr(t)k = (r(t) r(t)) = t (r(t) r(t)) = (r(t) r(t)) = (r(t) r 0 (t)) 5

6 i:e:; = r(t) r0 (t) (r(t) r(t)) = = r(t) r0 (t) kr(t)k t kr(t)k = kr0 (t)k cos ; = kr(t)kkr0 (t)k cos kr(t)k = kr 0 (t)k cos ; where is the angle between r(t) an r 0 (t). In general, t kr(t)k = kr0 (t)k cos 6= kr 0 (t)k = r(t) t : That is, t kr(t)k 6= r(t) t. The equality hols only when cos =, i.e., = n, n = 0; ; ; : : :, i.e., r(t) an r 0 (t) are parallel with pointing the same irection.. In the box, put V for a vector, S for a scalar an UD for an unene quantity. Give the reason for your answer. () (u v)w V Answer. u v is a scalar an the vector multiplie by a scalar is a vector. () kuk(v w) S Answer. The norm/magnitue of a vector is a scalar an the ot prouct prouces a scalar. Thus the expression is a scalar. () u v + w UD Answer. The ot prouct prouces a scalar. The sum of a scalar an a vector is not ene. (4) (v w) (u v) UD Answer. The ot prouct prouces a scalar. The cross prouct prouces a vector. The cross prouct between a scalar an a vector is not ene. (5) v w Answer. The expression vector is not ene. UD. () What is the object represente by the equation x + y = in the {imensional space? Justify your answer. 6

7 Answer. The equation represents a plane parallel to the z{axis. The plane has the x{intercept (; 0; 0) on the xy{plane an the y{intercept (0; ; 0) on the xy{plane. Since the plane is parallel to the z{axis, there is no z{intercept. Another way: The equation x + y = can be expresse as follows: (x 0) + (y 0) + 0(z c) = ; where c is any constant. This expression says that the object is a plane with the normal vector h; ; 0i an passing through the points (; 0; 0) an (0; ; 0). Please, confer the gure. Z Z Y Y. Graph of the plane x + y = for 0 x 0, 7 y an 0 z 0.. Graph of the line r(t) = t h; ; i for 00 t 00. () What is the object represente by the vector{value function r(t) = t i + t j + t k in the {imensional space? Justify your answer. Answer. We observe r(t) = D t ; t ; t E = t h; ; i : If we put various values of t an plot the vectors, we observe that it is a line with the irection h; ; i. Another way: Let x = t, y = t an z = t an let s = t. Then we have or x 0 x = s = 0 + s; y = s = 0 + s; z = s = 0 + s; = y 0 = z 0 = s; which is an equation of a line passing through the point (0; 0; 0) an parallel to the vector h; ; i. Please, confer the gure. 7