18.01 Final Answers. 1. (1a) By the product rule, (x 3 e x ) = 3x 2 e x + x 3 e x = e x (3x 2 + x 3 ). (1b) If f(x) = sin(2x), then

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1 8. Final Answers. (a) By the product rule, ( e ) = e + e = e ( + ). (b) If f() = sin(), then f (7) () = 8 cos() since: f () () = cos() f () () = 4 sin() f () () = 8 cos() f (4) () = 6 sin() f (5) () = cos() f (6) () = 64 sin() f (7) () = 8 cos(). (a) The line tangent to y = 5 + at = has a slope equal to that of the curve at = and passes through the point (, 4). The slope of the line at = is y ( = ) = 6 5 = 6() 5 = 7 = m. The y-intercept of the line, b, is found by using the slope and the known 4 b point: = 7 b =. The equation of the line is therefore y = m + b = 7. (b) If the curve had a horizontal tangent, then at some point the first derivative of y with respect to would be equal to zero. The derivative of the equation y + y = 4 is

2 y + (y )y + y + y = y (y + ) = y y. If y were equal to, then y y = y y =. This y + equation is valid when both and y are zero or when y = y for nonzero and y. The first case is not valid, because we are given that y + y = 4, which would not be possible if and y were both zero. The second case is also impossible, because y = y y = (we can divide by y because in this case it must be nonzero) and it is not possible for the ratio of two squares (necessarily positive numbers) to be equal to a negative number. Therefore y can never be zero and so the curve defined by y + y = 4 has no horizontal tangents.. (a) ( ) d f(t) f() = lim d + t t t = lim t+ + t t t( + ) (t + ) = lim t (t )(t + )( + ) t + t t = lim t (t )(t + )( + ) t = lim t (t )(t + )( + ) = lim t (t + )( + ) = ( + ) (b) tan () π/ lim When, the numerator becomes π/ π/ = and as the denominator also goes to zero, we can use l Hospital s rule to compute the limit:

3 (tan () π/) /( + ) lim = lim ( ) = lim + = + ( ) = 4 4. As shown in the graph below, y = has the following properties: + Local maimum (y =, y < ) at = Local minimum (y =, y > ) at =- The function is increasing (y > ) when < The function is decreasing (y < ) when > The inflection points (y = ) are =, ± The graph is symmetric about the origin ( ) The horizontal asymptote lim is the line y = + There is no vertical asymptote 5. The values and y are defined as in the figure below:

4 The area of printed type = 5 in, so y = 5 and the total area of the poster is ( + 4)(y + 8). To minimize the amount of paper used, we need to minimize the total area of the poster. ( + 4)(y + 8) = y + 4y = 8 + 4y + 8 since we know that y = 5. We can also substitute y = 5/, so that we have an area equal to: 4(5) To find the minimum of this equation we set the first derivative with respect to equal to zero: + 8 = = 5 = 5, taking only the positive root because represents a physical quantity. We can check that = 5 corresponds to a minimum of the area by taking the second derivative of +8, which is 4. Since this is positive at = 5, the point does indeed correspond to a minimum. If = 5 then y = 5 y =. Thus the dimensions of the poster which minimize the amount of paper used are a = + 4 = 9 in and b = y + 8 = 8 in. 6. Let y be the total distance from the plane to the car, and let be the horizontal distance between the plane and the car. The question asks for dc/dt, the car s speed. 4

5 From the Pythagorean theorem, y = +, because the plane is a distance one mile above the road. By definition, we also know that dc/dt = d/dt, as the plane has speed mph with respect to the ground. In addition, since y = / at t =, we know that = y = 5 at t =. We can then determine that: ( ) dy = ( + ) / d () = 6 dt dt and we can substitute = 5/ to obtain: ( ) 5 d = dt From this we can calculate: dc/dt = mph 5

6 7. (7a) n i ( ) lim + = + d n n n i= = ( + ) / = () / = (7b) lim sin( sin( ) )d = lim h h h h By l Hospital s rule, this is equal to 8. (8a) π/4 tan sec d = +h +h d lim sin(( + h) ) = sin(4) h π/4 ( ) sin π/4 sin d = d cos cos cos du Let u = cos. Then = sin(). Substituting into the integral, d π/4 =π/4 π/4 sin du d = = cos() ( = cos(π/4) =. cos = u ) (8b) Using integration by parts, ln d = ln d = (4) ln() ln() 4 = ln() ln() 4 9. Using the inverse trigonometric substitutions = sin θ, d = cos θdθ, the integral becomes 6

7 9 sin θ( cos θdθ) = sin θ sin θdθ. We can then use the double angle formula sin θ = ( cos θ) to obtain 9 ( cos θ) dθ. Evaluating the integral, we have 9 9 θ sin θ + C, 4 where C is a constant of integration. Substituting back in, d 9 ( ) = sin C *for reference, this is worked out in lec 5, fall 5, p.4. In general, the volume of an area revolved around the y-ais can be found by V = π b f()d a In this case, we are revolving the region as shown in the figure below: 7

8 Applying the formula to the region between a, a, = a, and = a/, we obtain: a V = π a d a/ Substituting u = and du/d = : =a ( ) V = π a udu = π (a u) / =a/ Replacing u with : a 4π V = ( ) (a ) / a/ 4π ( ) = ( a (a/ ) ) / ( ) 4π a / = 4 πa = =a =a/. Let y() = e. Using the two-trapezoid method, the picture should be approimately as follows: 8

9 The areas of the regions are then: Region I: ( )y() = y() = (.7) = 5.4 Region II: (5 )y() = y() = (6.7) =.4 Region III: (.5)( )(y() y()) = y() y() = = 4 Region IV:(.5)(5 )(y(5) y()) = y(5) y() = = And the total area is then 45.8 units.. (a) It is given that the rate of radioactive decay of a mass of Radium-6, dm/dt, is proportional to the amount m of Radium present at time t. We can then write dm = Am, dt where A is a constant. Re-writing and integrating the equation, dm = Adt m ln(m) = At + C m = e At+C = e At e C m = Ce At 9

10 where C is a constant. We can find A and C by using the information given in the problem. First, we know that there are mg of Radium present at t =, so that m(t = ) = C = mg. We also know that it takes 6 years for m to decrease by half. Therefore: (5/) =.5 = e 6A ln(.5) = 6A A = ln(.5)/6. Finally, m = Ce At = e (ln(.5)/6)t = (e ln(.5) ) t/6 = (.5) t/6, where t is in years and m(t) is in mg. (b) When t = years, and using the approimation given in the question, m = (.5) /6 = () /6 (.65) = 65mg.. The formula for arc length S of a curve defined by parametric equations (t) and y(t) is: In this problem, (t) is given as S = b (t) + y (t) dt. a t cos(πu /)du

11 and Their derivatives are y(t) = t sin(πu /)du. ( ) πt (t) = cos ( ) y πt (t) = sin Substituting (t), y (t), and the appropriate limits into the formula for arc length results in: S = t cos (πt /) + sin (πt /)dt = t dt t = t = t 4. (4a) The Taylor series of a function f() centered at = a is f (a)( a) f () (a)( a) f () (a)( a) f (4) (a)( a ) 4 f(a) !!! 4! The Taylor series of ln( + ) centered at = a is then ( + a) ( a) ( + a) ( a) ( + a) ( a) ()()( + a) 4 ( a) 4 ln(+a) !!! 4! And the Taylor series of ln( + ) centered at a = is therefore ()() 4 4 ln() = !!! 4! 4 = ( ) n= n+ n n (4b) Using the ratio test,

12 c n ( ) n+ n < = = n. c n+ ( ) n+ n + n + Because n is the inde of summation (an increasing integer), n + is always greater than n and therefore n < < n + Thus < and the radius of convergence is < <. (4c) ln(/) = ln( +.5) can be approimated by the first two non-zero terms of the Taylor series found in (a): ln( + ) +.5 =.5 = 8 (4d) The upper bound of the error in (c) s approimation is found using Taylor s inequality for an approimation of n terms: n+ R n () M n, (n + )! where = / and n =. In addition, M n f (n+) () M ( + ) for all /; the maimum of M in this range is for = /, which gives M = 6. Putting these numbers into the above formula, (.5) R n (.5) 6 =! 5. We can prove the inequality by showing that the derivatives of the terms satisfy the inequality for > and then by working backwards from there: ( ) d =, d(tan ()) =, d() = + + ( + ) +

13 < < for all > + ( + ) ( ) + t t t d < d < d for all > + ( + ) + t < tan (t) < t for all t > + t < tan () < for all > +

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