151. The position of a moving car is given by \( X(t) = 2t + 5 \) meters. What is the velocity of the car?
151. The position of a moving car is given by \( X(t) = 2t + 5 \) meters. What is the velocity of the car?
答案： ( C )
Velocity is the first derivative of position: \( X'(t) = 2 \) m/s.
152. A spherical balloon is being inflated. What is the volume of the sphere at the instant when the rate of increase of the surface area is four times the rate of increase of the radius?
答案： ( D )
\( S = 4\pi r^2 \); \( \frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 4\frac{dr}{dt} \) ⇒ \( r = \frac{1}{2\pi} \); \( V = \frac{4}{3}\pi r^3 = \frac{1}{6\pi^2} \) .
153. A conical funnel has base diameter 4 cm and height 5 cm. Water drains at \( 2\,\mathrm{cm}^3/\mathrm{s} \). How fast is the water level falling when the water is 2.5 cm high?
答案： ( D )
Similar triangles: \( r/h = 2/5 \); \( V = \frac{4}{75}\pi h^3 \); \( \frac{dV}{dt} = \frac{4}{25}\pi h^2 \frac{dh}{dt} \); with \( \frac{dV}{dt} = -2 \) and \( h = 2.5 \) get \( \frac{dh}{dt} = -\frac{2}{\pi}\) cm/s.
154. A light 50 m from a building shines on a 1-m tall child walking away at 1 m/s. How fast is the shadow on the building decreasing when the child is 10 m from the building?
答案： ( C )
\( S = 50/x \); \( \frac{dS}{dt} = -50 x^{-2} \frac{dx}{dt} \); when child is 10 m from building, \( x = 40 \); \( \frac{dS}{dt} = -\frac{50}{1600} = -\frac{1}{32}\) m/s.
155. A camera 100 m from a launch pad tracks a rocket rising at 20 m/s. How fast must the camera's angle change when the rocket is 100 m high?
答案： ( D )
\( \tan\theta = x/100 \); \( \sec^2\theta \frac{d\theta}{dt} = \frac{1}{100}\frac{dx}{dt} \); at \( x = 100 \), \( \sec\theta = \sqrt{2} \); \( \frac{d\theta}{dt} = \frac{1}{10}\) rad/s ≈ 5.7 deg/s.
156. The width of a rectangle increases twice as fast as the length. How does \( \frac{dA}{dt} \) compare to \( \frac{dw}{dt} \) when the length is twice the width?
答案： ( D )
\( A = lw \), \( l = 2w \), \( \frac{dw}{dt} = 2\frac{dl}{dt} \); \( \frac{dA}{dt} = 2w\frac{dw}{dt} + w\cdot\frac{1}{2}\frac{dw}{dt} = 2.5w\frac{dw}{dt} \) .
157. Given \( C(x) = 200 + 4x + 0.5x^2 \), what is the marginal cost when \( x = 50 \)?
答案： ( B )
Marginal cost \( C'(x) = 4 + x \); \( C'(50) = 54 \) .
158. Position \( s(t) = t^3 - 6t^2 + 9t \). At what times is the velocity zero?
答案： ( B )
\( v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3) = 0 \) at \( t = 1, 3 \) s.
159. What does \( \lim_{h\to 0} \frac{v(3+h) - v(3)}{h} \) mean?
答案： ( C )
Acceleration is the derivative of velocity; this limit is \( a(3) \) .
160. A man and woman leave an intersection at 1 m/s, east and south. When each is 100 m from the intersection, how fast is the distance between them increasing?
答案： ( B )
\( z^2 = x^2 + y^2 \); \( 2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \); at \( x = y = 100 \), \( z = 100\sqrt{2} \); \( \frac{dz}{dt} = \frac{2}{\sqrt{2}} \approx 1.4 \) m/s.
161. What is the instantaneous rate of change of \( f(x) = 2x^2 - 4x + 2 \) when \( x = 20 \)?
答案： ( B )
\( f'(x) = 4x - 4 \); \( f'(20) = 76 \) .
162. Position \( s(t) = 2t^3 - 4t^2 + 2t - 5 \) (meters). What is the instantaneous acceleration at \( t = 10 \)?
答案： ( A )
\( a(t) = s''(t) = 12t - 8 \); \( a(10) = 112 \) m/s².
163. Cost \( C(x) = 200 + 16x + 0.1x^2 \), computers sell for $500 each. If 1,000 are produced and sold, what is the marginal profit?
答案： ( D )
\( P = 500x - C(x) \); \( P'(x) = 484 - 0.2x \); \( P'(1000) = 284 \) .
164. A right circular cone has fixed base radius 4 cm. Height increases at 0.5 cm/min. What is the rate of change of volume?
答案： ( D )
\( V = \frac{1}{3}\pi r^2 h \); \( \frac{dV}{dt} = \frac{1}{3}\pi r^2 \frac{dh}{dt} = \frac{1}{3}\pi (16)(0.5) = \frac{8\pi}{3} \approx 8.4 \) cm³/min.
165. What are the coordinates of the vertex of \( y = 3x^2 + 2 \)?
答案： ( D )
\( y' = 6x = 0 \) at \( x = 0 \); \( y(0) = 2 \); vertex (0,2).
166. BC: Water pumped into a cylindrical tank (r=10 m) at 100 m³/min and removed at 70 m³/min. Approximate rate of change of water level?
答案： ( C )
\( \frac{dV}{dt} = 100 - 70 = 30 \); \( V = \pi r^2 h \) ⇒ \( \frac{dh}{dt} = \frac{30}{100\pi} \approx 0.1 \) m/min.
167. An 8-ft ladder slides down the wall at 1 ft/s. How fast is the bottom sliding away when the top is 4 ft from the ground?
答案： ( C )
\( x^2 + y^2 = 64 \); \( x\frac{dx}{dt} + y\frac{dy}{dt} = 0 \); at \( y = 4 \), \( x = 4\sqrt{3} \); \( \frac{dx}{dt} = -\frac{y}{x}\frac{dy}{dt} = \frac{\sqrt{3}}{3}\) ft/s.
168. 1-kg iron block; \( Q = mc\Delta T \), c=460 J/(kg·K). If \( \frac{dQ}{dt} = 100 \) J/s, approximate rate of temperature change?
答案： ( B )
\( \frac{dQ}{dt} = mc\frac{d(\Delta T)}{dt} \) ⇒ \( \frac{d(\Delta T)}{dt} = \frac{100}{460} \approx 0.22 \) K/s.
169. A balloon is inflated at 50 cm³/s. How fast is the diameter increasing when the radius is 10 cm?
答案： ( D )
\( V = \frac{4}{3}\pi r^3 \); \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \); \( D = 2r \) ⇒ \( \frac{dD}{dt} = \frac{dV/dt}{2\pi r^2} = \frac{50}{200\pi} = \frac{1}{4\pi}\) cm/s.
170. A farmer has 160 m of fence to enclose a rectangular area against a straight river (three sides). What is the maximum area?
答案： ( D )
\( 2x + y = 160 \), \( A = xy = x(160-2x) \); \( \frac{dA}{dx} = 160 - 4x = 0 \) ⇒ \( x = 40 \), \( y = 80 \); \( A = 3200 \) m².
171. Given \( C(x) = 144 + 0.1x + 0.04x^2 \), what is the minimum average cost per unit?
答案： ( C )
\( \bar{C} = C/x = 144/x + 0.1 + 0.04x \); \( \bar{C}' = -144/x^2 + 0.04 = 0 \) ⇒ \( x^2 = 3600 \), \( x = 60 \); min average cost $60.
172. \( G(t) \) gives the rate a rain gauge is filling (mm/min). What does \( G'(4) \) mean?
答案： ( A )
\( G(t) \) is a rate; \( G'(t) \) is the rate of change of that rate (mm/min per min).
173. Find two negative numbers that add to -50 with maximum product.
答案： ( D )
\( P = x(-50-x) = -50x - x^2 \); \( P' = -50 - 2x = 0 \) ⇒ \( x = -25 \), both -25.
174. At some instant the volume of a sphere is \( 1/(48\pi^2) \) cubic units. How does the rate of increase of surface area compare to the rate of increase of the radius?
答案： ( D )
\( V = \frac{4}{3}\pi r^3 \) ⇒ \( r = 1/(4\pi) \); \( S = 4\pi r^2 \) ⇒ \( \frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 2\frac{dr}{dt} \) .
175. Open box from 4 cm × 8 cm cardboard by cutting squares from corners. What square length gives the greatest volume?
答案： ( C )
\( V = x(4-2x)(8-2x) \); \( V' = 0 \) gives \( x \approx 0.845 \) cm (other root out of range).
176. Cost \( C(x) = 500 + 6x + 0.2x^2 \), price \( p(x) = 20 \). How many units for maximum profit?
答案： ( D )
\( P = 20x - C(x) = -500 + 14x - 0.2x^2 \); \( P' = 14 - 0.4x = 0 \) ⇒ \( x = 35 \) .
177. Find two positive numbers that add to 30 with maximum product.
答案： ( D )
\( P = x(30-x) = 30x - x^2 \); \( P' = 30 - 2x = 0 \) ⇒ \( x = 15 \) .
178. Which illustrate the Chain Rule? I. \( \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} \) II. \( y = f(g(x)) \) ⇒ \( \frac{dy}{dx} = f'(g(x))g'(x) \) III. \( y = f(u) \) ⇒ \( \frac{dy}{dx} = f'(u)\frac{du}{dx} \) IV. \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
答案： ( D )
All four are valid statements of the Chain Rule or related differentiation.
179. 1-kg copper block; \( Q = mc\Delta T \), c=390 J/(kg·K). If \( \frac{d(\Delta T)}{dt} = 0.026 \) K/s, approximate rate of thermal energy added?
答案： ( A )
\( \frac{dQ}{dt} = mc\frac{d(\Delta T)}{dt} = 1(390)(0.026) \approx 10 \) J/s; closest is 1.0 (answer key gives 10 J/s, option A).
180. Piano on 90-ft rope over pulley 40 ft above man's arms. At t=0 man is 30 ft from piano and walks away at 12 ft/s. How fast is the piano being pulled up?
答案： ( D )
Geometry: \( x^2 + 40^2 = (50+h)^2 \); at t=0, \( 30^2+40^2 = (50+h)^2 \) ⇒ \( h=0 \); \( \frac{dh}{dt} = \frac{x}{50+h}\frac{dx}{dt} = \frac{30}{50}(12) = 7.2 \) ft/s.
181. Position \( s(t) = t^3 - 9t^2 + 24t \). At what time is the acceleration zero?
答案： ( D )
\( a(t) = s''(t) = 6t - 18 = 0 \) ⇒ \( t = 3 \) s.
182. A 16-in piece of metal is used for a gutter; two sides of length \( x \) are turned up. How many inches should be turned up for maximum capacity?
答案： ( D )
\( A = x(16-2x) \); \( A' = 16 - 4x = 0 \) ⇒ \( x = 4 \) in.
183. Ellipsoid \( r_a=8 \) cm (constant), \( r_b=4 \) cm, \( r_c=2 \) cm; \( \frac{dr_b}{dt}=0.5 \) cm/min, \( \frac{dr_c}{dt}=2 \) cm/min. \( V = \frac{4}{3}\pi r_a r_b r_c \). Initial rate of change of volume?
答案： ( D )
\( \frac{dV}{dt} = \frac{4}{3}\pi r_a (r_c \frac{dr_b}{dt} + r_b \frac{dr_c}{dt}) = \frac{32\pi}{3}(2(0.5)+4(2)) = 96\pi \) cm³/min.
184. A plane lifts off at 30° at 500 mi/h. How fast is it gaining altitude?
答案： ( D )
\( \frac{dh}{dt} = \sin(30°) \cdot 500 = 250 \) mi/h.
185. Cost \( C(x) = 300 + 4x + 0.2x^2 \), product sells for $500 each. What is the marginal profit function?
答案： ( D )
\( P = 500x - C(x) \); \( P'(x) = 500 - 4 - 0.4x = 496 - 0.4x \) .
186. 600 m of fence to enclose a rectangular area against a river (three sides). Maximum area?
答案： ( D )
\( 2x + y = 600 \), \( A = x(600-2x) \); \( x = 150 \), \( y = 300 \); \( A = 45{,}000 \) m².
187. Oil pumped into cylindrical tank (r=100 m) at 5,000 m³/min and removed at 100 m³/min. Approximate rate of change of oil level?
答案： ( D )
\( \frac{dV}{dt} = 5000 - 100 = 4900 \); \( \frac{dh}{dt} = \frac{4900}{\pi(100)^2} \approx 0.16 \) m/min.
188. Position \( s(t) = 16t^3 - 14t^2 + 6t - 7 \) (meters). Instantaneous velocity at \( t = 0.5 \)?
答案： ( C )
\( v(t) = 48t^2 - 28t + 6 \); \( v(0.5) = 12 - 14 + 6 = 4 \) m/s.
189. What is the maximum volume of a cylinder inscribed in a cone with altitude 15 cm and base radius 3 cm?
答案： ( D )
Similar triangles: \( h = 5(3-r) \); \( V = \pi r^2 h = 5\pi r^2(3-r) \); \( \frac{dV}{dr} = 0 \) ⇒ \( r = 2 \), \( h = 5 \); \( V = 20\pi \) cm³.
190. Pressure \( \Delta p = 2\gamma/r \). When \( \frac{dS}{dt} = 4\left(-\frac{dr}{dt}\right) \) (γ constant), how does \( \frac{d(\Delta p)}{dt} \) compare to \( \frac{dr}{dt} \)?
答案： ( A )
\( S = 4\pi r^2 \) ⇒ \( 8\pi r \frac{dr}{dt} = 4\frac{dr}{dt} \) so \( r = 1/(2\pi) \); \( \frac{d(\Delta p)}{dt} = -2\gamma r^{-2}\frac{dr}{dt} = -8\pi^2\gamma \frac{dr}{dt} \) .
191. A cylindrical can (base diameter 10 cm, height 10 cm) is full. Water drains at 4 cm³/s. How fast is the water level falling?
答案： ( C )
\( V = \pi r^2 h \), \( r = 5 \) constant; \( \frac{dh}{dt} = \frac{1}{\pi r^2}\frac{dV}{dt} = \frac{-4}{25\pi}\) cm/s.
192. Given \( C(x) = x^2 + 20x + 4 \), what is \( x \) so that the average cost is minimum?
答案： ( A )
\( \bar{C} = x + 20 + 4/x \); \( \bar{C}' = 1 - 4/x^2 = 0 \) ⇒ \( x = 2 \) .
193. Velocity \( v = \frac{1}{3}t^3 - 2t^2 + 3t + 2 \). At what time(s) does acceleration equal zero?
答案： ( D )
\( a = \frac{dv}{dt} = t^2 - 4t + 3 = (t-1)(t-3) = 0 \) at \( t = 1, 3 \) s.
194. Football height \( y(t) = v_0 t - \frac{1}{2}gt^2 \), \( v_0 = 20 \) m/s, \( g = 10 \) m/s². How long to reach maximum height?
答案： ( D )
\( y' = v_0 - gt = 0 \) ⇒ \( t = v_0/g = 20/10 = 2 \) s.
195. Position \( s(t) = 2t^2 - 20t + 5 \) (meters). What is the object's acceleration?
答案： ( A )
\( a(t) = s''(t) = 4 \) m/s² (constant).
196A. BC 196. Rocket altitude \( y(t) = \frac{1}{3}t^3 - 2t^2 + 3t + 2 \) (m). (A) What is the velocity function (m/s) and the acceleration function (m/s²)?
Velocity: \( y'(t) = t^2 - 4t + 3 \) m/s. Acceleration: \( y''(t) = 2t - 4 \) m/s².
196B. 196. (B) Make graphs of all three functions vs. time for 0 to 5 s. Derive the maxima and minima for each graph.
Altitude: max at (1, 3.3), min at (3, 2). Velocity: max at endpoints, min at (2, -1). Acceleration: linear from -4 to 4; zero at t=2.
197A. 197. Conical funnel (base 6 cm, height 5 cm) over cylindrical can (diameter 4 cm, height 5 cm). Water drains at 2 cm³/s. (A) How fast is the water level in the funnel falling when the water is 2.5 cm high?
\( V = \frac{3}{25}\pi h^3 \); \( \frac{dV}{dt} = \frac{9}{25}\pi h^2 \frac{dh}{dt} \); \( -2 = \frac{9}{25}\pi (2.5)^2 \frac{dh}{dt} \) ⇒ \( \frac{dh}{dt} = -\frac{8}{9\pi} \approx 0.28 \) cm/s.
197B. 197. (B) How fast is the water level in the can rising?
\( \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \) with \( r = 2 \) and inflow 2 cm³/s ⇒ \( \frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} \approx 0.16 \) cm/s.
197C. 197. (C) Will the can overflow? If not, how high will the final water level be?
Funnel volume \( \frac{1}{3}\pi(3)^2(5) = 15\pi \) cm³; can volume \( \pi(2)^2(5) = 20\pi \) cm³. Can does not overflow. Final height \( h = \frac{15\pi}{\pi(2)^2} = 3.75 \) cm.
198A. BC 198. Isosceles triangle has perimeter 16 cm. (A) Make a drawing of the problem.
Draw triangle with two equal sides \( x \), base \( 16 - 2x \); height from vertex to base.
198B. 198. (B) What are the dimensions of the sides and height for the maximum area?
\( A = (8-x)(2\sqrt{x-4}) \); \( \frac{dA}{dx} = 0 \) ⇒ \( x = 16/3 \) cm, base = 16/3 cm, height = \( \frac{8\sqrt{3}}{3} \) cm; triangle is equilateral.
198C. 198. (C) What is the maximum area?
\( A = \frac{1}{2}(16/3)\cdot \frac{8\sqrt{3}}{3} = \frac{32}{3}\sqrt{4/3} \approx 12.3 \) cm².
199A. 199. Cost \( C(x) = 50 + 2x + x^2/20 \), revenue \( R(x) = 20x + x^2/200 \). (A) Derive the weekly profit function.
\( P(x) = R(x) - C(x) = 20x + \frac{x^2}{200} - 50 - 2x - \frac{x^2}{20} = 18x - 50 - \frac{9x^2}{200} \) .
199B. 199. (B) What is the maximum weekly profit?
\( P'(x) = 18 - 0.09x = 0 \) ⇒ \( x = 200 \) shirts; \( P(200) = 3600 - 50 - 1800 = \$1750 \) .
200A. 200. Make a cylindrical tin can with closed top to hold 360 cm³. (A) Make a drawing and label what is given.
Draw cylinder with radius \( r \), height \( h \); given \( V = 360 \) cm³.
200B. 200. (B) What are its dimensions if the amount of tin used is to be minimum?
\( V = \pi r^2 h = 360 \) ⇒ \( h = 360/(\pi r^2) \); \( S = 2\pi rh + 2\pi r^2 \); \( \frac{dS}{dr} = 0 \) ⇒ \( r^3 = 180/\pi \), \( r \approx 3.86 \) cm, \( h \approx 7.71 \) cm.
200C. 200. (C) What is the surface area?
\( S = 2\pi r h + 2\pi r^2 \approx 187 + 94 = 281 \) cm².