MAC 2311 Calculus I Spring 2004

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1 MAC 2 Calculus I Spring 2004 Homework # Some Solutions.#. Since f (x) = d dx (ln x) =, the linearization at a = is x L(x) = f() + f ()(x ) = ln + (x ) = x. The answer is L(x) = x..#4. Since e 0 =, and d dx ex = e 0 =, x=0 the linearization of e x at 0 is, in fact e x e 0 + e 0 (x 0) = + x. To determine the values of x for which it is accurate to within 0. we have to use some calculating device. If we compare the graphs of y = e x and y = x +, we see that one always has e x > + x, except at 0 where it is equal. One also sees there will be two values of x, one negative (call it x ), one positive (call it x 2 ) at which e x = + x The interval [x, x 2 ] is then exactly the set of all values of x for which e x is within 0. of + x. To find x, x 2 we must solve e x = +x+0.. Using a calculator to solve this equation I got the (approximate) roots 0.482, The answer is: For x #0. The sketch will be omitted, but is provided upon request in my office. If y = /x, x = 4, and dx = δx =, then y = y(x + x) y(x) = y() y(4) = 4 = 4 ; dy = dy dy (x)dx = (4)dx = dx dx x 2 ( ) =. x=4.# is the value at 99.8 of the function f(x) = x. The point nearest to 99.8 where we know all about he square root is x = 00. We thus use: f(99.8) f(00)+f (00)( ) = ( ) = = #8. If we consider the function f(x) = x, then (if we accept that linear approximations are reasonable), since f (x) = x 5, (x + x) x + x 5 x.

2 For x =, x = 0.0 this gives (.0) + (0.0) =.0. Chapter, Review Section #95. Drawing a picture helps! The idea is that the sun is far away enough from the earth so that its rays can be considered as parallel lines. At least, locally. The picture below shows the east side of the building as the vertical line on the left. One of the sun s rays is pictured as a slanted line. The variable x represents the length of the shadow and θ is the angle of elevation of the sun, which is the angle between the slanted line (ray of the sun) and the horizontal line (ground). h x θ The relation between x and θ is h/x = tan θ where h = 400 (feet0 is the height of the building; i.e., x tan θ = 400. Differentiating both sides of this identity with respect to t gives dx dt tan θ + x sec2 θ dθ dt = 0. We have to solve for dx/dt when θ = π/. We are given that dθ/dt = 0.25 (radians per hour; negative because it is decreasing). All we are still missing is the value of x at the time when the angle is π/, which is easily computed from the relation x tan θ = 400. We recall that tan π/ = /, cos π/ = /2, sec 2 π/ = 4/. Solving, we get first that x = 400 (feet), then dx 4 dt = 400 ( 0.25) = 400 (feet/hour). Chapter, Review Section #9. (a) L(x) = 4 4 (x ) = 4 x (b) Omitted. (c) 2.29 < x <.47. Problems Plus, #4. The beam of light from the headlights of the car is a ray tangent to the parabola. From the picture, we see that the equation of the parabola must be of the form y = ax 2 with a > 0. It being given that the car starts at the point ( 00, 00), we should have 00 = a( 00) 2 from which 2

3 a = /00 = 0.0. The equation of the parabola is thus y = 0.0x 2. It follows that when the car is at position (x 0, y 0 ) the equation of the tangent line is T (x) = y 0 + 2(0.0)x 0 (x x 0 ). If we consider that y 0 = 0.0x 2 0 we can rewrite this in the form T (x) = 0.0x x 0 x. Because the car moves in a positive x direction, the light beam is the half of this line where x x 0. It is now easy to figure out when the statue will be illuminated. The statue is at (00, 50) so we need to find a value of x 0 so that the point (00, 50) is on the line of equation T (x) given above with x > x 0. Or: We need to find x 0 so that if T (x) is as given above, there is x > x 0 with (x, T (x)) = (00, 50). Thus x = 00 and 0.0x x 0 x = 50; i.e., since x = 00, 0.0x 2 0 2x = 0. We solve this quadratic equation for x 0 ; the two solutions are x 0 = 00 ± Because we need x 0 < x = 00, the x 0 value we select is the one with the minus sign; x 0 = The answer is that the headlights of the car illuminate the statue when the car is at the point of coordinates ( , ) (29.289, 8.579). Problems Plus, #7. This exercise can be done without using any calculus. A circle with center on the axis of the parabola will intersect the parabola in 4 points, 2 points or no points. If it has exactly two points of intersection, then it is tangent to the parabola. We seek the center of a circle of radius, center on the positive y-axis, which has exactly two points of intersection with the parabola of equation y = x 2. The equation of a circle of radius center at (0, b) (b > 0) is x 2 + (y b) 2 =, or x 2 + y 2 2by + b 2 = 0. If it intersects the parabola at (x, y), then y = x 2 and we get the following equation for x: x 2 + x 4 2bx 2 + b 2 = 0 or x 4 + ( 2b)x 2 + (b 2 ) = 0. This is a quadratic equation in x 2 with the solutions x 2 = 2b ± 5 4b. 2

4 Now (think a bit about it) there should be exactly positive solution of this equation. Each positive solution produces two possible values of x (x = ± x 2 ) so if we have more than one, we have too many. Less than positive solution gives us no intersection at all. All this leads us to conclude that we must have b = 5/4. The center should be placed at (0, 5 ) = (0,.25). 4 Solution 2, with Calculus. Geometry shows us that the line containing the radius of the circle from the center to the right part of the parabola has to be perpendicular to the tangent to the parabola at that point. If the point of contact circle-parabola is (x, y) = (x, x 2 ) and the center is at (b, 0), then we get the equations: x 2 + (x 2 b) 2 = 2 = (Radius of the circle is ), b x 2 x = 2x (Radius is perpendicular to tangent line). From the second one we get x 2 = b 2 ; using this in the first one we get once again b = 5/4. Determine all the critical points of x(x 2 ). If we analyze where the expression inside the absolute values is positive and where it is negative, we get the following description of the function: f(x) = x(x 2 ) = x + x, x <, x(x 2 ) = x x, x < 0, x(x 2 ) = x + x, 0 x <, x(x 2 ) = x x, x. Thus, at first glance, x 2 +, x <, f x (x) = 2, < x < 0, x 2 +, 0 < x <, x 2, < x. By looking at the left and right limits at, 0,, we see that we cannot improve any of the strict inequalities to a non-strict one; f is NOT differentiable at, 0,. This makes them critical points. Away from these points, the derivative is zero if x = ±/. The critical points are,,, 0,,. 4.#48. We have f (x) = x 2. Setting to 0 produces the roots x = ±, of which only is in the interval. Since f(0) =, f() =, f() = 9, 4

5 the absolute maximum value is 9 (not (, 9)), the absolute minimum value is. 4.#0. We have f (x) = ( ln x)/x 2 which is zero precisely at x = e, which is in the interval in question. We have: f() = 0, f(e) = ln, f() = e. The minimum is clearly 0. The maximum value is whichever one of /e and ln / is larger. A calculator settles the score; the larger one (by a hair) is /e. 4.#(b). We have f (x) = (x 2 )e x x which is zero precisely at x = ±/. Of these, only is in the relevant interval. We have: f( ) =, f( ) = e 2 /9 >, f(0) =. The minimum value is. The maximum value is e 2 /9. 4.#78. (a) The critical numbers of the given cubical function are precisely the roots of ax 2 + 2bx + c = 0, a quadratic equation. numbers. Example: If 4b 2 2ac > 0, then there are exactly two critical f(x) = x + x 2. If 4b 2 2ac = 0, then there is exactly one critical number. Example: f(x) = x. If 4b 2 2ac < 0, then there is no critical number. Example: f(x) = x + x. (b) A cubic function can have either exactly two local extreme values (a local max and a local min) or none. 4.2#8. Let f(x) = 2x sin x. This is a continuous function. Since f(0) = and f(π) = 2π > 0, it must have a zero in the interval [0, π], by the Intermediate Value Theorem. Thus the equation f(x) = 0 has at least one root. If it had two (or more roots), since between any two zeros of a function there is a zero of its derivative, the equation f (x) = 2 cos x = 0 would have at least one root. But this is impossible, since cos x for all x. 4.2#20. If the equation had at least three real roots, the derivative of f(x) = x 4 + 4x + c would have at least two zero. But f (x) = 4x + 4 = 0 has a single solution, namely x =. 4.2#24. By the mean value theorem there is some c (2, 8) such that f(8) f(2) = f (c). 5

6 We can use now that f (c) 5 to get f(8) f(2) 5; that is, 8 f(8) f(2) #4. If v(t) is velocity as a function of time, then acceleration is a(t) = v (t). We have to be careful here with the units! Measuring time in hours so that 2:0 becomes 2 =, the mean value theorem tells us that there is some time c between 2 : 00 and 2 : 0 such that a(c) = v (c) = v( ) v(2) = 50 0 = #4. The maximum value has to be assumed at a critical number. We have f (x) = ae bx2 + 2abx 2 e bx2 = 0 if x 2 = /2b. If this is to happen for x = 2 we need b = /8. More precisely, 2 is a critical point if and only if b = /8. Before we wonder if it is a maximum or not, we notice that to get f(2) = now requires 2ae /2 =, hence a = e /2 /2. It is now easy to see that there is a maximum at x = 2. The answer is a = 2 e 2, b = 8.

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