Math 1A: Homework 6 Solutions

Transcription

1 Math A: Homework Solutions July 30. Sketch the graphs of the following functions. Ensure that your work includes: Domain Intercepts Symmetry Asymptotes and End-behaviour Intervals of Increase/Decrease/Local Max/Min Concavity and Inflection Points a) fx) x)e x. We have: Domain: R. Intercepts: we have f0) 0)e 0 so the y-intercept is at y. For the x-intercept, set fx) 0 to get x)e x 0 x 0 x. Symmetry: f x) + x)e x so f doesn t have any obvious symmetry. Asymptotes and End-behaviour: note that as x, x) and e x so lim x)e x. We also have lim x x)e x e x 0. e x Intervals of Increase/Decrease/Local Max/Min: we have f x) x) e x ) e x e x x ). For x <, f x) < 0; for x >, f x) > 0 and f x) 0 at x. Concavity and Inflection Points: we have f x) e x ) e x x ) e x 3 x). For x < 3, f is concave up since f x) > 0; for x > 3, f is concave down since f x) < 0. As f ) > 0, we also conclude that f has a relative minimum at x. )

2 Figure Gluing this together, we get Figure. b) fx) x 8 x. We have: Domain: since 8 x 0 x 8 x 8. The domain is therefore [ 8, 8]. Intercepts: f0) 0 so the y-intercept is at y 0. For the x-intercept, set fx) 0 to get x 8 x 0 x 0, ± 8. Symmetry: since f x) x 8 x fx), f is an odd function. Asymptotes and End-behaviour: f has no horizontal or vertical asymptotes as the domain is bounded and f does not blow up at any point. Intervals of Increase/Decrease/Local Max/Min: we have f x) 8 x + x x) 8 x x 8 x 8 x. 8 x 8 x Note then that at 8 x 0 x ±, f x) 0; when x <, f x) > 0 and when x >, f x) < 0. Concavity and Inflection Points: we have 8 x 4x) 8 x ) x f ) 8 x x) 8 x 4x8 x ) + x8 x ) 8 x ) 3/ x3 4x 8 x ) 3/ xx ) 8 x ) 3/. Note then that f ) < 0 and f ) > 0 so f has relative maximum minimum) at x x ). Moreover, f x) < 0 for 0 < x < 8 f is concave down on these x-values.

3 Gluing this together, we get Figure c) fx) tan x x+). We have Figure Domain: recall that tan ) is defined for all real inputs. We only need to ensure that x exists; hence, the domain is R { }, ), ). x+ Intercepts: since f0) tan ) π/4, the y-intercept is at y π/4. Next, set fx) 0 to get ) ) x x tan 0 0 x. x + x + Symmetry: since f x) tan x x+), there is no obvious symmetry. Asymptotes and End-behaviour: Note that as x ±, x x+) ; thus, lim x ± fx) tan ) π/4. Also observe that as x +, x x+) so limx + fx) y tan y) π/. Likewise, as x, x x+) so limx fx) y tan y) π/. Intervals of Increase/Decrease/Local Max/Min: we have f x) + x x+ ) d dx x + ) x + ) + x ) ) x x + ) x + )) x )) x + ) ) x + ) x + ) + x ) x + ) x + ) + x ). 3

4 Observe that f x) > 0 for all x. Concavity and Inflection Points: we have f x) d x + ) + x ) ) dx x + ) + x ) ) x + ) + x )) x + ) + x ) ) 8x x + ) + x ) ). Hence, f x) > 0 whenever x < 0 and f x) < 0 whenever x > 0. We therefore have the graph in Figure Figure 3 d) fx) x sinx) for π x 3π. We have Domain: [ π, 3π]. Intercepts: f0) 0 so the y-intercept is y 0. Set fx) 0 to get x sinx) 0. One solution to this equation is at x 0; moreover, by drawing rough sketches of y x and y sinx), it can be seen that the equation is also satisfied for some x in π/, π) and x in π, π/). Symmetry: since f x) x sin x) x + sinx) fx), f is odd. Asymptotes and End-behaviour: since the domain is bounded and f does not blow up at any point, we do not need to check for asymptotes. Intervals of Increase/Decrease/Local Max/Min: we have f x) cosx). 4

5 Set f x) 0 to get cosx) / x ±π/3, ±5π/3, 7π/3. We can also infer that when x [ π, 5π/3) π/3, π/3) 5π/3, 7π/3) f x) < 0 x 5π/3, pi/3) π/3, 5π/3) 7π/3, 3π] f x) > 0. Concavity and Inflection Points: we have f x) sinx) so f is concave up when x is in π, π) 0, π) π, 3π) and concave down when x is in π, 0) π, π). We therefore have the curve shown in Figure Figure 4 e) fx) lnsinx)). We have Domain: ln ) is defined only when the argument is positive; hence we require sinx) > 0 x... π, π) 0, π) π, 3π) 4π, 5π).... Intercepts: x 0 does not belong to the domain so f has no y-intercept. For the x-intercept, we require fx) 0 lnsinx)) 0 sinx) x π/ + kπ where k is an integer. Symmetry: observe that if sinx) > 0, then sin x) sinx) < 0 and vice versa; this shows that f is defined either at some x and not x or the other way about. Hence, f does not have even-odd symmetry. Note however that fx + π) lnsinx + π)) lnsinx)) fx) so f is periodic with period π. This ensures that we only need to sketch the function on 0, π) and draw similar curves at distances of π, 4π,.... Asymptotes and End-behaviour: note that as x 0 +, sinx) 0 + so lnsinx)). Similar behaviour occurs when x π/. 5

6 Intervals of Increase/Decrease/Local Max/Min: we have f x) cosx) sinx) cotx). We deduce that x 0, π/) f x) > 0, x π/, π) f x) < 0 and f π/) 0. Concavity and Inflection Points: we have f x) csc x) which is always negative. Hence f is always concave down. We end up with the sketch in Figure Figure 5 f) fx) x + x x. We have Domain: we require x + x 0 xx + ) 0 x, x 0. The domain therefore is, ] [0, ). Intercepts: we have f0) 0 so the y-intercept is at y 0. For the x- intercept, set fx) 0 x + x x x + x x x 0. Symmetry: since f x) x x + x, there is not obvious symmetry. Asymptotes and End-behaviour: Note that as x, x + x and x) ) so fx). Also note that if x > 0, we can write fx) x + x. As x, we get an indeterminate form of the type.0 so we need to use

7 L Hopital s Rule: lim x + x ) + x x + x x + x x ). Intervals of Increase/Decrease/Local Max/Min: we have f x) x + x + x. Set f x) 0 to get x + x + x x + x + x x + ) 4x + x) 4x + 4x + 4x + 4x 0. Since this is impossible, this shows that f x) is never zero. Since f is continuous wherever defined, it must take the same sign on the same branch. For example, f ) 5 < 0 so f x) < 0 for all x <. Similarly, since f ) 3 > 0, f x) > 0 for all x > 0. Concavity and Inflection Points: we have f x) x + x) x + ) x+ x +x ) 4x + x) 4x + x) x + ) 4x + x) 3/ 4x + 4x) 4x + 4x + ) 4x + x) 3/ 4x + x). 3/ This shows that f x) < 0 always so f is concave down throughout. We end up with the curve shown in Figure.. Evaluate the following limits. 7

8 Figure a) lim x 0 x x. Note that this is of the form 0 0 so we instead consider lim x 0 lnxx ) x 0 x lnx) lnx) x 0 x x x 0 x 3 x x 0 0. / so L Hopital) It follows that lim x 0 x x e 0. b) lim x 0 + sinx) lnx). Note that this is of the form 0. ) so we need to rewrite this as a quotient. We have by L Hopital s Rule lnx) lim sinx) lnx) x 0 + x 0 + cscx) /x lim x 0 + cscx) cotx) / ) x) x 0 + x cosx) sinx) cosx) x 0 + x sinx) + cosx) 0/0) x

9 c) lim x +7 3x cos 5x). We have lim x + 7 3x cos 5x) x + 7)/x 3x cos 5x))/x since cos 5x) x 0 as x. d) lim x 0 sinhx) x x 3. We have by L Hopital s Rule + 7 x 3 cos 5x) x 3 sinhx) x coshx) lim 0/0) x 0 x 3 x 0 3x x 0 sinhx) x x 0 coshx). 0/0) 0/0) e) lim + a bx. x) Note that this is of the form if b > 0) or if b < 0). We therefore consider lim ln + a ) bx b lim x ln + a ) x ) x ln + a x b lim b lim b lim ab. x + x) a ax ) x a + a x It follows that lim + a x) bx e ab. ) 0/0 so L Hopital) 3. Find the dimensions of the lightest open-top cylindrical can that will hold a volume of 000 cm 3. Let r and h represent the radius and height in cm) respectively of the can. We are given that πr h 000 h 000. πr We need to minimize the surface area ) 000 A πr + πrh πr + πr πr 9 πr r

10 where r > 0. We have da dr 000 πr. Set this equal to zero to get r πr 000 r r π r 0 π /3. Since d A π > 0 when r 0/π /3, we conclude that A has a relative dr r 3 minimum at r 0/π /3. Since this is the only critical point, we conclude that A also has an absolute minimum at this point. When the radius is 0/π /3 cm, the corresponding height is h 000 π00/π /3 ) 0/π/3 cm. 4. The U.S. Postal Service USPS) accepts a parcel for shipment only if the sum of its length and girth distance around the middle) does not exceed 75 cm. What dimensions will give a square-ended-box the largest possible volume? Let l, w, h be the length, width and height of a box. Since the box is square ended, we infer that w h. We also require that the length plus girth not exceed 75 cm; for the largest box, we should stretch this allowance to the maximum; we therefore have We need to maximize the volume l + w + h 75 l + 4w 75 l 75 4w. V lwh lw 75 4w)w. Since w is a length, we need it to be nonnegative; at the same time, we need l 0 so w 75/ Hence, we need to maximize V 75 4w)w over [0, 8.75]. For the critical points, we have: the end-points w 0, we have dv 75 4w)w) dw 4w 550w w. Set this equal to zero to get w 0 and w V is differentiable everywhere on 0, 8.75). Note that V 0) 0, V 8.75) 0 and V 45.83) 953. cm 3 so the volume is maximized when w 550/ cm. The corresponding length is l 75/3 9.7 cm. 5. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed at m. Find the proportions of the window that will admit the most light. Let the length of the window be x m and the height be y m. The diameter of the semicircle then is x as well. The perimeter of the window is therefore given by y + x + πx y 3 x + π ). 0

11 Suppose that the clear glass admits K units of light per m ; the tinted glass then admits K/ units of light per m. The total light admitted is given by L Kxy) + K π x ) ) [ K x 3 x + π )) ] + πx. Note that we need x, y 0 3 x maximize F x 3 x + π For the critical points, we have the end-points x 0, + π ). note that Set df dx ) + π 0 x )) + πx over [0, + π ) ]. df 3 dx x + π )) 3 x + π ) 3 x 3πx 8. 0 to get x 3 +3π/8. F is differentiable at all points of Observe that F 0) 0, F + π ) ) π + πx 8 ) 0,. + π ) + π ) + π ) + x + π ) ) + πx 8 ).095 and F Hence, the light through the window is maximized when x 3 y.9 m.. We therefore need to +3π/8 3 +3π/8 ) m and. The coastline in Torquay runs east-to-west. Basil, a coastguard 00 m north of the coastline, spots a swimmer Manuel in need of assistance. At that time, Manuel is 400 m south-east of Basil. Given that Basil can run at 8 m/s and swim at 5 m/s, chart the most efficient course for Basil. Let x be the horizontal distance covered on land see Figure 7). The distance covered on land then is d L x Since the original horizontal and vertical distances from Basil to Manuel are both 400/ 00, Basil still needs to swim 00 x) horizontally and 00 00) vertically. Hence, the distance to be swum is d W 00 00) + 00 x). The total time for such a course is T x) d L 8 + d W 5 x ) + 00 x) +. 5

12 Basil x 00 m 400 m Manuel Figure 7 Observe that even though x can take any real value, we can restrict its possible values to [0, 00 ] since otherwise the coastguard will be retracing his steps. Next, note that T x) x x x 00 ) ) x) Set T x) 0 to get x x + 00 x 00 ) ) + 00 x) 5x 00 00) + 00 x) 8x 00 ) x x 00 00) + 00 x) ) 4x 00 ) x + 00 ). This yields a quartic equation whose two real solutions are x 4.95 m and x.75 m. Since the former is outside our domain, we ignore it. Note that T 0) ) +00 ) s. 8 5 T ) ) s. 8 5 T.75) 7. s.

13 Hence, the coastguard should cover.75 m horizontally before entering the water. In other words, he should go tan ) east of south before entering the water. 7. a) Let fx) x + for x > 0. Find the absolute minimum of f. x We have fx) x + so f x). Set f x) 0 to get x x ±. x x Since only x > 0 is allowed, we only consider x. Also note that f x) so f ) > 0 so f has a relative minimum at x 3 x. Since this is the only stationary point, we conclude that f has an absolute minimum of f) at this point. b) Prove that if a, b, c are positive real numbers, then + a ) + b ) + c ) 8abc. Since a, b, c > 0, from a) we can infer that a + a b + b c + c Multiply these to get ) ) ) a + b + c + a b c 3 + a ) + b ) + c ) 8abc. 8. When we cough, the trachea windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity v in cm/s) can be modeled by the equation v cr 0 r)r, r 0 r r 0 where r 0 is the rest radius of the trachea in centimeters and c is a positive constant whose value depends in part on the length of the trachea. Show that v is greatest when r 3 r 0. Since we are finding the absolute maximum of v when r takes values in a closed and bounded interval, we only need to consider the end-points r r 0 /, r 0. 3

14 for the derivative, we have dv dr c )r + r 0 r)r)) crr 0 3r ) crr 0 3r). Set this equal to zero to get r 0 or r 3 r 0. Note that the former does not belong to the domain. no points come from the non-differentiability of v. We then have vr 0 /) cr 0 /)r 0 /) cr 3 0/8, vr 0 ) 0 and vr 0 /3) cr 0 /3)r 0 /3) 4cr 3 0/7. Note that since 4 7 > 8, v takes it largest value at r 3 r 0. 4