Mathematics 115 Professor Alan H. Stein April 18, 005 SOLUTIONS 1. Define what is meant by an antiderivative or indefinite integral of a function f(x). Solution: An antiderivative or indefinite integral f(x) dx is a function whose derivative is f(x).. An object starts moving along a straight line from rest. Its acceleration at time t is equal to 1 + cos t, where t is measured in seconds and its distance is measured in feet. (a) Find its equations of motion, that is, find formulas for its velocity and distance travelled. Use appropriate notation and clearly indicate what each variable you use represents. Solution: Let v represent the velocity and s the distance. We know dv dt = 1 + cos t, so v = 1 + cos t dt = t + sin t + k for some constant k. Since the object starts at rest, v = 0 when t = 0, so we have 0 = 0 + sin 0 + k = k and v = t + sin t. We know ds dt = v = t + sin t, so s = t + sin t dt = t cos t + c for some constant c. Since s = 0 when t = 0, we get 0 = 0 cos 0 + c = 0 1 + c = c 1, so c = 1 and s = t cos t + 1. (b) How fast is it going after π seconds? Solution: v t=π = π + sin π = π + 0 = π, so the object is moving at a speed of π feet per second after π seconds. (c) How far does it go in π seconds? Solution: s t=π = π π cos π + 1 = + π feet in π seconds. π ( 1) + 1 = +, so the object travels Page 1 of 5
3. Consider the function f(x) = x 8 ln x. Page of 5 (a) Analyze monotonicity, that is, determine where the function is increasing and where it is decreasing. Solution: f (x) = x 8/x = (x 4) (x + )(x ) =. x x Since f is only defined for x > 0 (because ln is only defined on R + ), there is just one critical point, x =. Clearly, f (x) > 0 for x > while f (x) < 0 for 0 < x <, so f is decreasing on (0, ) and increasing on (, ). (b) Find and identify all local and global extrema. Solution: Clearly, f has a local and global minimum at but has no maxima of any kind. Extra Credit: Analyze concavity and sketch its graph. Solution: f (x) = + 8/x. Clearly, f (x) > 0 for all x in the domain of f, so the function is concave up everywhere. Note that the y axis is a vertical asymptote, since lim x 0 + =. 4. Calculate x (x + 4) 3 dx. Solution: Use the substitution u = x + 4. We have du du = x, du = xdx, dx = dx x. Plugging into the original integral, we get x (x + 4) dx = x 3 u du 3 x = 1 1 u du = 1 3 u 3 du = 1 u = 1 1 1 = 4 u 4(x + 4) +k.
Page 3 of 5 5. Use a Tangent Line Approximation involving the function f(x) = 3 x to estimate 3 64.1. Solution: Writing f(x) = x 1/3, we get f (x) = 1 3 x /3 = 1 3x = 1 /3 3 3. We thus have x f(64) = 3 64 = 4 and f (64) = 1 3 3 64 = 1 3 4 = 1 48. Using the Tangent Line Approximation, we get T (x) = f(x 0 ) + f (x 0 )(x x 0 ) = 4 + 1 (x 64). 48 We thus have 3 64.1 T (64.1) = 4 + 1 0.1 (64.1 64) = 4 + 48 48 = 4 + 1 480 4.0008333. 6. Use Newton s Method on the function f(x) = x 3 64.1 with x 0 = 4 to estimate 3 64.1. Calculate x 1 and x. Solution: f (x) = 3x. Using Newton s Method, we have x n = x n 1 f(x n 1) f (x n 1 ) = x n 1 x3 n 1 64.1. 3x n 1 Letting x 0 = 4, we have: x 1 = 4 43 64.1 3 4 4.0008333333. x = 4.0008333333 4.00083333333 64.1 3 4.0008333333 4.0008491.
7. Find the point on the line y = x + 3 closest to the point (0, 13). Page 4 of 5 Solution: Let z be the square of the distance between a point (x, y) on the line and the point (0, 13). z will be minimal for the point closest to the point (0, 13). Using the distance formula, z = (x 0) + (y 13) = x + (y 13). We may proceed three different ways. (a) Since the point is on the line y = x + 3, we have z = x + (x + 3 13) = x + (x 10). The minimum must occur at a point where z = 0. We calculate z = x + (x 10) = x + 8x 40 = 10x 40 = 10(x 4). Clearly, z = 0 when x = 4, at which point y = 4 + 3 = 11. So the point on the line closest to (0, 13) is (4, 11). (b) Since y = x + 3, we have dy dx =. Also, since z = x + (y 13), we obtain dz dz = x + (y 13)dy. Since the minimal value of z must occur when = 0, we dx dx dx may solve the following equations simultaneously: y = x + 3 dy dx = x + (y 13) dy dx = 0. Using the fact that dy = and dividing both sides of the third equation by, dx we obtain x + (y 13) = 0, x + y 6 = 0. Plugging in y = x + 3, we get x + (x + 3) 6 = 0, 5x 0 = 0, 5x = 0, x = 4. As before, we get y = 4 + = 11 and determine the closest point is (4, 11). (c) This question can also be answered without the use of Calculus, since it is known that the line containing the closest point and the point (0, 13) is perpendicular to the given line. The given line has slope, so the perpendicular line has slope 1 and equation y 13 = 1(x 0) or y 13 = 1 x. Since the point of intersection is also on the line y = x + 3, we may plug in y = x + 3 to get (x + 3) 13 = 1 x, x 10 = 1 x, (x 10) = x, 4x 0 = x, 5x = 0, x = 4. We then get y = 4 + 3 = 11, so the point we want is (4, 11)
Page 5 of 5 8. As the sun sets, its angle of inclination is decreasing at a rate of 0. radians per hour. How fast is the shadow of a foot tall tree lengthening when the angle of inclination of the sun is 30 or π/6 radians? The angle of inclination is the angle between the ground and a line going from a point on the ground to the sun. Solution: Let x be the length of the shadow and let θ be the angle of inclination. We know tan θ = dθ dx and = 0. and want to find θ = π/6. x dt dt Differentiating, we get d dt (tan θ) = d ( ) dθ, sec θ dθ dt x dt = dx dx x sec θ, so x dt dt = dt. 3 When θ = π/6, we have cos θ =, so sec θ =, so sec θ = 4 3 3. We also have tan θ = 1, so 3 x = 1 so x = 3. 3 We thus obtain dx ( 3) 4 dt = 3 ( 0.) = 40, so the shadow is increasing at a rate of 40 feet per hour.