2011 Spring Math 151
Exam I Version B Solutions
9. E The angle between the gravity force (weight) and the motion of the block is 60 , so the work done is W = |F||D| cos 60 = 1 = 300 ft-lbs. (30)(20) 2 10. B f (3) = lim f (3 + h) - f (3) = lim h2 - 4h + 7 = 7. h0 h The equation of the line whose slope is 7 and which passes through the point (3, 4) is y - 4 = 7(x - 3), or y = 7x - 17.
h0
1. C Set x and y components equal and solve for t. t = 4 at (2, -3), t = 0 at (0, 1), t = 1 at (1, 0), but -1 = t has no solution, so the point not on the curve is (-1, 0). 2. A cos x 1 - sin x cos x(1 + sin x) 1 - sin2 x sec x + tan x. (1 + sin x) (1 + sin x) 1 + sin x = cos x = =
11. E lim
x(x - 6) = lim x = 6. x6 x - 6 x6
-1 cos x 1 3. A Since -1 cos x 1, x x x 1 Since 0 as x , by the Squeeze x Theorem, the limit is 0. 4. B As we approach 4 from the left along the curve, the y-value approaches 2. 5. B Let f (x) = x3 - x2 + x. f is continuous since it is a polynomial, and f (2) = 6, f (3) = 21, so f (2) < 10 < f (3). Therefore, by the Intermediate Value Theorem, there is a solution to f (x) = 10 on [2, 3]. 6. C a = (6i + 3j) - (3i - j) = 3i + 4j. To form ^ a unit vector a, multiply by the reciprocal 1 ^ (3i + 4j) = of the magnitude: a = 2 + 42 3 3 4 i + j. 5 5 7. D
x5+ x5-
12. E (a) is both continuous and differentiable everywhere; (c) and (d) are not continuous at x = 0. From the graph of (e), we see that the function continuous but not differentiable at x = 0 is 3 x
Graph of x(1/3) 1.5
1
0.5
0
-0.5
-1
-1.5 -2
-1.5
-1
-0.5
0
0.5
1
1.5
2
x5
f (5) = 2. Therefore, since lim+ f (x) = f (5) = lim- f (x), f is continx5
lim f (x) = -8 + 2(5) = 2,
lim f (x) = 6 5 - = 1.
2 + 3x - 5 2 + 3x + 5 f (x) - f (1) = lim 13. f (1) = lim x1 h0 x-1 x-1 2 + 3x + 5 2 + 3x - 5 = lim x1 (x - 1)( 2 + 3x + 5) 3(x - 1) = = lim x1 (x - 1)( 2 + 3x + 5) 3 3 = . lim x1 2 + 3x + 5 2 5 14. . (a) projn m = = 4+2 mn n = 4, 2 2 |n| (42 + 22 ) 1, 1 -
and
uous only from the right. 8. B Factor x from numerator and denom8 x(3 - ) 3 x inator: lim = . Similarly, 9 x 4 x(4 + ) x 8 x(3 - ) x = 3. lim 9 x- 4 x(4 + ) x 1
6 3 , so orthn m = , 5 5 1 2 6 3 = - , . , 5 5 5 5
(b) The point (0, 0) is on the line, so let b = 1, 1 - 0, 0 = 1, 1 . The vector v = 4, 2 is in the direction of the line, so v = -2, 4 is orthogonal to the line. The distance is found
2 = 20 + (NOTE that this is the magnitude of the vector in (a) above!) by |compv b| = (-2)2 42
3 2.5
-2 + 4
2
n
1.5
1
m
0.5 0
orthn m
projn m
-0.5 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
15. Find
a common denominator: 2 1 x - 2 2 x = lim 1 2 - x lim x2 x - 2 x2 x-2 2x 1 1 =- . = lim - x2 2x 4 16. F1 = 8 cos 45 , 8 sin 45 = 4 2, 42 . F2 = 14 cos 60 , -14 sin 60 = 7, -7 3 . The resultant force is F1 + F2 = Therefore, 4 2 + 7, 4 2 - 7 3 . the magnitude of the resultant force is (4 2 + 7)2 + (4 2 - 7 3)2 .
1 x
17. We use r(t) = r0 +tv to find the parametric equations. r0 corresponds to the point at t = 1 : r(1) = 5i + 5j. v = ((5t)i + (8 - 3t2 )j) - (5i + 5j) = r (1) = lim t1 t-1 (5t - 5)i + (3 - 3t2 )j 5t - 5 lim i+ = lim t1 t1 t-1 t-1 5(t - 1) 3(1 + t)(1 - t) lim i + lim j t1 t - 1 t1 t-1 = 5i - 6j. Therefore, the vector equation of the tangent line is r(t) = (5i + 5j) + t(5i - 6j), so the parametric equations are r(t) = (5 + 5t)i + (5 - 6t)j, or x = 5 + 5t, y = 5 - 6t.
3 - 3t2 t-1
j=
2