Solutions to Test #1 MATH 2421

Transcription

1 Solutions to Test # MATH Pulhalskii/Kawai (#) Decide whether the following properties are TRUE or FALSE for arbitrary vectors a; b; and c: Circle your answer. [Remember, TRUE means that the statement is true for ANY combination of vectors a; b; and c:] (a) T F: a b b a: FALSE. Cross product is generally NOT commutative. In general, we have (b a) a b: (b) T F: a (b + c) (a b) + (a c) TRUE. Dot product can be distributed over addition. Thus, it acts like multiplication. (c) T F: (a) b a (b) : TRUE. Multiplicative scalars can be factored in front of everything. (a) b (a b) a (b) (a b) : (#) Let a i + 5j k and b i 7j + k: Find the engineering form of a b: The determinant is i j k 5 7 i 5 7 j + k 5 7 i (0 ) j ( ) + k ( ( 5)) i j 9k: x y (#) The limit lim (x;y)!(0;0) + y does not exist. However, if we choose the path y x ; as x! 0; we obtain a real number. What is that number? Fractions, fractions... x x lim x!0 lim lim lim x!0 + + x!0 + x!0 : x (#) The shape of a hill is given by the equation z f (x; y) x y + 50; where f gives us the altitude in feet. The highest point of the hill occurs when x 0 and y 0: At the point P (; ; ) ; we want to nd which direction (in D) will water naturally ow down the hill from that point P: Hint: It s the direction which has the lowest value of D u f (; ) :

2 The gradient vector will give us the direction for the steepest ascent (going up!). So the antigradient (opposite) direction will give the direction of steepest descent. Water will naturally ow down the hill in this direction. rf hf x ; f y i x ; y rf (; ) ; h ; 96i : We want the opposite direction, so the correct vector is h ; 96i h; 96i : (#5) The equation x y z 9 gives us a quadric surface. (a) What is the name of this surface? This was a standard form with two negative signs. HYPERBOLOID OF TWO SHEETS. (b) Sketch the trace for z 0: If we set z 0; we have x y ; a hyperbola. y The asymptotes are y x : x (#6) Suppose we have a force vector F h; 7i and a displacement vector! P Q h5; i. (a) Find the work accomplished by this force when an object is displaced over! W ork F P Q h; 7i h5; i 0 + ( 7) work units. (b) Use the note sheet and evaluate Proj! P Q F: (#7) NOT a perfect square. Proj! P Q F!! F P Q! h; 7i h5;!! P Q P Q P Q i h5; i h5; i 5 h5; i h5; i : + ( ) 6 h5; i! P Q: Suppose we have a position function when 0 t : Find the arc length of the associated parametric curve. r (t) sin (t) ; cos (t) ; t Hint: Your nal answer should be in the form n where n is a positive integer.

3 We need the speed function. v (t) cos (t) ; sin (t) ; t Dcos (t) ; sin (t) ; t E kv (t)k qcos (t) + ( sin (t)) + t q cos (t) + sin (t) + t p t + : (#8) Here are two lines which intersect at some point in D. L : L : x t + x s + y t y s z t z s + (a) Find values for t and s at the point of intersection and write down the coordinates of the point of intersection P: Set the coordinates equal to each other. t + s + t s t s + There should be a pair (s; t) which satis es all three equations. The last equation gives us t s : We substitute this into the rst equation. ( s ) + s + s + s + s s s 0: If we check all three equations, we end up with t : The coordinates (x; y; z) are (; ; ) : (b) Find the angle of intersection for the two lines. We can read o the coe cients of t in order to nd the direction vector associated with that line. For L ; we have u h; ; i and for L ; we have v h; ; i : The dot product gives us cos () Thus, we have cos (0) 90 : h; ; i h; ; i kh; ; ik kh; ; ik 6 p p 0: 9 7 (#9) If f (x; y) ln (x + y) ; evaluate f xy at the point (; ) : f [ln (x + x + [x + x + y (x + y) : [f h (x + y) i (x @y [x + y] (x + y) () (x + y) :

4 (#0) Match the equations below with their associated surfaces. Assume that the surface presentations have the positive x-axis coming out of the paper, and the positive y-axis pointing to the right, etc. (i) (D) z p x + y : Right circular cone (concave downward). (ii) (E) (iii) (B) z x + y : Elliptic paraboloid. x 9 + y : Elliptic cylinder. (iv) (F) z x y : Hyperbolic paraboloid (saddle shape). (#) Evaluate the directional derivative D u f (; 0) if u is the unit direction vector associated with h; The unit vector is i and f (x; y) h; i q p h; i : + ( ) 0 x x + y : The gradient vector is rf (x + y) (x) x () hf x ; f y i (x + y) ; x (x + y) () x + xy (x + y) ; x (x + y) : Thus, we have D u f (; 0) rf (; 0) p 0 h; i + () (0) ( + 0) ; ( + 0) p h; 0 i p (h; i h; i) p ( + ) p p 0 : (#) Suppose we have a D surface Evaluate the implicit derivative We have and z cos (y + z) + x cos at the point P (; 0; 0). F (x; y; z) z cos (y + z) + x cos (y) 0; evaluated at P (; 0; 0) : Fx F z F x x F z z ( sin (y + z)) + cos (y + z)

5 Thus, Fx F z x z sin (y + z) + cos (y + z) : 0 + cos (0) (#) Here is the position function for projectile motion (launched from the point P (0; y 0 ): r (t) v 0 cos () t; 6t + v 0 sin () t + y 0 : A cannonball is launched from the top of a building y 0 96 feet. The initial speed is 6 p ft/sec, and 5 is the angle of elevation above the horizon. (a) Find the time t when the cannonball hits the ground. We choose t seconds. 6t + 6 p sin (5 ) t t + 6t t t 6 0 ) t ; : (b) Based on the information given to you in this problem, give the algebraic expression for the unit tangent vector function, T (t) : (The general function do NOT evaluate it for a particular value of t:) SIMPLIFY as much as possible. We have v (t) hv 0 cos () ; t + v 0 sin ()i h6; t + 6i : T (t) v (t) h6; t + 6i : kv (t)k q6 + ( t + 6) It turns out that we can factor out (6) from the numerator and denomi- This is okay. nator. T (t) h; t + i q + ( t + ) : (#) Use the total di erential to estimate the change z when x moves from 0.0 to 0. and y moves from :00 to :98: z f (x; y) y e x The total di erential is dz f x dx + f y dy y e x dx + ye x dy and we have x 0; dx +0:; y ; and dy 0:0: dz e (0) (+0:) + () e (0) ( 0:0) : 0:08 +:: The change in z will be approximately +:: 5

6 (#5) Suppose w f (x; y; t) x x (s; t) y y (t) : After substituting, w is really a function of s and t. Write the appropriate Multivariable Chain : f. # & x y t. # # s t t There are three paths of change, and thus, three sets of @y + : (#6) Extra Credit. Find the antiderivative. sec () d??? Integration by parts: u sec () ) du tan () sec () d dv sec () d ) v tan () This gives us sec () d tan () sec () tan () tan () sec () d tan () sec () tan () sec () d: Now use the trig. identity: tan () sec () : sec () d tan () sec () tan () sec () tan () sec () sec () sec () sec () + sec () d: sec () d sec () d: We transfer the R sec () to the left side. sec () d tan () sec () + sec () d: 6

7 If you memorized this, then you probably completed everything! sec () d ln jtan () + sec ()j + C: Now divide both sides by. Thus, we have sec () d (tan () sec () + ln jtan () + sec ()j) + C: 7