Math 3 Exam Spring 06 Name: ID: 7 multiple choice questions worth 4.7 points each. hand graded questions worth 0 points each. 0. free points (so the total will be 00). Exam covers sections.7 through 3.0 No graphing calculators! Any non-graphing, non-differentiating, non-integrating scientific calculator is fine. For the multiple choice questions, mark your answer on the answer card. Show all your work for the written problems. Your ability to make your solution clear will be part of the grade. sin(a ± B) sin A cos B ± sin B cos A tan(a ± B) tan A ± tan B tan A tan B sin (A/) cos A sin A sin B [cos(a B) cos(a + B)] cos sin A cos B [sin(a + B) + cos(a B)] ( ) ( ) A + B A B sin A + sin B sin cos ( ) ( ) A + B A B cos A + cos B cos cos Law of Cos: c a + b ab cos C d ( sin x ) dx x d ( tan x ) dx + x cos(a ± B) cos A cos B sin A sin B tan(a/) cos A sin A sin A + cos A cos (A/) + cos A A cos B [cos(a B) + cos(a + B)] ( A + B sin A sin B cos ( A + B cos A cos B sin ) sin ) sin Law of Sin: sin A sin B sin C a b c d ( cos x ) dx x ( ) A B ( ) A B
Math 3 Exam Page of 7. Let f(x) x x 3 +. Find f () A. B. 0 C. /6 D. /8 E. /4 F. / G. H. 3/ I. Solution: Quotient rule f (x) x(x3 + ) (x )(3x ) (x 3 + ) x x4 (x 3 + ) f () 4. Let f(x) ln(x 3). Find f () A. 0 B. ln C. D. E. e F. 3 G. 4 H. 5
Math 3 Exam Page 3 of 7 Solution: Chain rule f (x) x x 3 f () 4 4 3. Find d dt (t5 + 3t e 4 ) /3 A. (t 5 + 3t e 4 ) /3 B. C. D. E. F. 3(t 5 + 3t e 4 ) /3 5t 4 + 6t (t 5 + 3t e 4 ) /3 5t 4 + 6t 4e 3 (t 5 + 3t e 4 ) /3 (5t 4 + 6t) 3(t 5 + 3t e 4 ) /3 π e 3 Solution: Chain rule d dt (t5 + 3t e 4 ) /3 3 (t5 + 3t e 4 ) /3 (5t 4 + 6t) (5t 4 + 6t) (t 5 + 3t e 4 ) /3
Math 3 Exam Page 4 of 7 4. Find d tan x dx A. B. sin x cos 3 x sin x cos3 x C. D. E. cos x sin x cos3 x sin x sin x cos x F. sec x Solution: d d tan x (tan x)/ dx dx (tan x) / (tan x) (tan x) / (sec x) (cos x) / (sin x) / (cos x) (sin x) / (cos x) 3/ sin x cos 3 x 5. Find d dx (ex sin 3x) A. e x ( sin 3x + 3 cos 3x) B. e x (3 sin 3x + cos 3x) C. e x ( sin 3x 3 cos 3x) D. e x (3 sin 3x cos 3x) E. e x sin 3x F. 3e x cos 3x G. 5e x cos 3x
Math 3 Exam Page 5 of 7 Solution: (e x sin 3x) (e x ) (sin 3x) + (e x )(sin 3x) e x (sin 3x) + 3e x cos 3x e x ( sin 3x + 3 cos 3x) 6. Let f(x) e (x3 x ). Find f () A. 0 B. C. D. e E. 3 F. 4 G. e Solution: f (x) (3x 4x)e x3 x f () ( 8)e 0 4
Math 3 Exam Page 6 of 7 7. Let y mx + b be the tangent line to the graph of f(x) 3x at the point (, ). What is m + b? A. B. 0 C. D. E. 3 F. 4 G. 5 H. 6 Solution: f (x) 6x, f () 6. Thus the equation of the tangent line is y 6x 5. 8. At the point (, ), the parabola y ax + bx has a tangent line equal to y 3x. Find a b. A. 3 B. C. D. 0 E. F. G. 3 H. π/3
Math 3 Exam Page 7 of 7 Solution: Set up equations to solve for a and b. Since the point (, ) is on the graph, we can plug this in and get the equation a + b. We also know that slope of the tangent line at x is 3. Since y ax + b, we can plug in y 3 and x into this: a + b 3 a + b Solving gives a and b.
Math 3 Exam Page 8 of 7 9. If g() 3 and g (), find d ( g(x) ) at x. dx x A. 5/4 B. C. 3/4 D. /4 E. 0 F. /4 G. 3/4 H. I. 5/4 Solution: d g(x) dx x x g (x) g(x) x d g(x) dx x g () g() x ( ) (3) 4 5 4 0. Find the value of c such that the line y 3 x + 6 is tangent to the curve y c x. A. It is not possible to find such a value of c. B. C. D. 3 E. 6 F. 9 G. 5 H. 36
Math 3 Exam Page 9 of 7 Solution: Taking the derivative: y c. We don t know the point that the x tangency occurs, but lets say it is at x a and y b. We have several conditions: 3 c a (Because y (a) 3/) b 3 a + 6 b c a We have three equations and three unknowns, we now solve (using your favorite method) to arrive at a 36/9, b, and c 6.
Math 3 Exam Page 0 of 7. Let f, g, h be differentiable functions. Here is a table of values for the functions and their derivatives. x f(x) f (x) g(x) g (x) h(x) h (x) 0 0 0 3 3 / 0 3 0 0 / /4 3 / 0 3 3 0 Find (f g h) (). A. -3 B. - C. - D. 0 E. F. G. 3 Solution: (f g h) () f (g(h())) g (h()) h () f (g(0)) g (0) () f ( ) (/) () (/) (). If g(x) + x cos(g(x)) x, find g (0). A. It is impossible to determine g (0). B. C. D. /
Math 3 Exam Page of 7 E. 0 F. G. Solution: Take the derivative (implicitly) (g(x)) + (x cos(g(x))) (x ) g (x) + cos(g(x)) xg (x) sin(g(x)) x Plug in x 0: g (0) + cos(g(0)) + 0 0 g (0) cos(g(0)) We now need to find g(0). We plug in x 0 into the original equation g(0) + 0 0 So g(0) 0. Thus, g (0) cos(0).
Math 3 Exam Page of 7 3. If f(x) ln(x + ln x), find f (). A. 0 B. C. D. e E. 3 F. e G. 0 H. e 3 Solution: f (x) x + ln x (x + ln x) f () ( + 0) ( + ) x + ln x x x + x(x + ln x) 4. Find the equation of the tangent line to the curve x xy y at the point (, ). A. y x B. y x + 3 4 C. y x 3 4 D. y x + 3 4 E. y 3 4 x + F. y 3 4 x G. y 3 4 x + Solution: Take the derivative (implicitly): (x ) (xy) (y ) () x (y + xy ) yy 0 y x y x + y Plug in (x, y) (, ) to get y (, ) 3/4. Thus the equation of the tangent line is y + 3 (x ). 4
Math 3 Exam Page 3 of 7 5. Let L(x) be the linearization of f(x) x at the point x 9. Find L(8). A. 3 B. 3 3 C. 3 6 D. 3 9 E. 3 8 F. 3 G. 3 + 8 H. 3 + 9 I. 3 + 6 J. 3 + 3 K. 3 + Solution: We have the formula for the tangent line: L(x) f(a) + f (a)(x a). Using f(x) x and a 9 gives. L(x) 3 + (x 9) 6 Plugging in x 8 gives L(8) 3 6.
Math 3 Exam Page 4 of 7 6. Find y by implicit differentiation when sin y x A. cos y B. sin y x C. sin y D. sin y x + E. sin y cos y F. G. H. sin y cos y sin y cos 3 y sin y cos 3 y Solution: Take the derivative implicitly, twice (sin y) (x) () (cos y)y 0 y sec y y (sec y tan y)(y ) (sec y tan y)(sec y) sec y tan y sin y cos 3 y 7. Suppose you have a function such that f(x + h) f(x) h cos h + h x + 3hx Find f ( ). A. It is impossible to determine f ( ).
Math 3 Exam Page 5 of 7 B. C. D. 0 E. F. G. 00 H. h Solution: f f(x + h) f(x) (x) lim h 0 h h cos h + h x + 3hx lim h 0 h h(cos h + hx + 3x) lim h 0 h lim cos h + hx + 3x + 3x h 0 Thus, f ( ).
Math 3 Exam Page 6 of 7 Name: ID: Written Problem. You will be graded on the readability of your work. Use the back of this sheet, if necessary. 8. Use the limit definition of derivative to find the derivative of f(x) x + Solution: f f(x + h) f(x) (x) lim h 0 h lim h 0 x+h+ x+ h x + x + h + lim h 0 h x + h + x + ( ) x + x + h + lim h 0 h x + h + x + (x + ) (x + h + ) lim h 0 h x + h + x + ( x + + x + h + ) lim h 0 h ( ) x + + x + h + x + + x + h + h x + h + x + ( x + + x + h + ) lim h 0 x + h + x + ( x + + x + h + ) x + x + ( x + + x + ) (x + )( x + ) (x + ) (x + 3/ )3/
Math 3 Exam Page 7 of 7 Name: ID: Written Problem. You will be graded on the readability of your work. Use the back of this sheet, if necessary. 9. On a dark night, a 6 foot tall person is walking away from a lamp post at a rate of 0 feet per second. The lamp post is 0 feet high. When the person is 5 feet from the lamp post, at what rate is the person s shadow growing or shrinking? Solution: Start with a picture. 0 feet 6 feet x y We know dx dy 0. We want to find dt dt do this, we use similar triangles: when x 5. We need to relate x and y. To 0 x + y 6 y We simplify this to get the equation: 0y 3x + 3y or, even better, 7y 3x. We now take derivatives and get 7 dy 3 dx dx dy. Plugging in 0 gives 30. dt dt dt dt 7