201. A function \( y \) has a tangent at \( x = a \) and a parallel tangent at \( x = b \). Which statement must be true?
201. A function \( y \) has a tangent at \( x = a \) and a parallel tangent at \( x = b \). Which statement must be true?
答案： ( B )
Parallel tangents have equal slopes, so \( \frac{dy}{dx} \) at \( x=a \) equals \( \frac{dy}{dx} \) at \( x=b \).
202. Velocity \( v(t) = t^3 - 12t^2 + 5 \). What is the particle's instantaneous velocity at \( t = 3 \)?
答案： ( B )
\( v(3) = 27 - 108 + 5 = -76 \) .
203. What is the approximate value of \( \sqrt[4]{78} \) using linear approximation?
答案： ( D )
\( f(x)=x^{1/4} \), \( a=81 \), \( \Delta x=-3 \); \( f(78)\approx f(81)+f'(81)(-3) = 3 + \frac{1}{4}(81)^{-3/4}(-3) \approx 2.97 \) .
204. BC: Slope of the line tangent to \( r = 1 + \sin\theta \) at \( (1+\sqrt{3}/2, \pi/3) \)?
答案： ( C )
\( \frac{dy}{dx} = \frac{(dr/d\theta)\sin\theta + r\cos\theta}{(dr/d\theta)\cos\theta - r\sin\theta} \); at \( \theta=\pi/3 \) with \( dr/d\theta=\cos\theta=1/2 \) gives \( \frac{1+\sqrt{3}}{1-\sqrt{3}} \) .
205. BC: Velocity vector \( \langle 3t^2, -9t \rangle \). Magnitude of acceleration at \( t=3 \) to nearest tenth?
答案： ( C )
\( \mathbf{a} = \langle 6t, -9 \rangle \); at \( t=3 \): \( |\mathbf{a}| = \sqrt{324+81} \approx 20.1 \) .
206. Penny dropped from 200 ft; \( s(t) = -16t^2 + 200 \), \( t\geq 0 \). To the nearest second, when does it hit the ground?
答案： ( B )
\( s(t)=0 \) ⇒ \( -16t^2+200=0 \) ⇒ \( t = \sqrt{200/16} = \frac{10\sqrt{2}}{4} \approx 3.5 \) → 4 s.
207. Tangent line to the curve at \( (x,y) \) is \( x - 3y = 1 \). What is the slope of the normal line?
答案： ( A )
Tangent slope \( = 1/3 \); normal slope \( = -1/(1/3) = -3 \) .
208. \( f \) differentiable, \( f(5)=10 \), \( f'(5)=2 \). Approximate \( f(5.5) \)?
答案： ( C )
Linear approximation: \( f(5.5)\approx f(5)+f'(5)(0.5) = 10 + 2(0.5) = 11 \) .
209. BC: Position \( \langle 2t^2-1, 4t \rangle \). Speed at \( t=1 \)?
答案： ( D )
\( \frac{dx}{dt}=4t=4 \), \( \frac{dy}{dt}=4 \); speed \( = \sqrt{16+16} = 4\sqrt{2} \) .
210. Position \( s(t) = t^3 - 9t^2 + 24t - 2 \). During what time interval is the particle moving in the negative direction?
答案： ( B )
\( v(t)=3t^2-18t+24=3(t-2)(t-4)=0 \) at \( t=2,4 \); \( v(3)<0 \) so negative on \( (2,4) \) .
211. Approximate \( \cos 62° \) using linear approximation.
答案： ( B )
\( f(x)=\cos x \), \( a=\pi/3 \) (60°), \( \Delta x = \pi/90 \); \( f(62°)\approx \cos(\pi/3) - \sin(\pi/3)(\pi/90) \approx 0.47 \) .
212. At \( x=a \), \( \frac{dy}{dx} \) does not exist but \( \frac{dx}{dy}=0 \). Describe the tangent(s) to \( y \) at \( x=a \).
答案： ( C )
\( \frac{dx}{dy}=0 \) implies vertical tangent (infinite slope \( dy/dx \)).
213. Approximate \( (5.2)^3 \) using linear approximation.
答案： ( C )
\( f(x)=x^3 \), \( a=5 \), \( \Delta x=0.2 \); \( f(5.2)\approx 125 + 3(25)(0.2) = 140 \) .
214. Velocity \( v(t) = 2t^3 - \frac{1}{2}t^2 + 4t - 6 \). Acceleration at \( t=4 \)?
答案： ( B )
\( a(t)=v'(t)=6t^2-t+4 \); \( a(4)=96-4+4=96 \) .
215. BC: Motion \( y=-2t^2+5t \), \( x=6t \). Speed at \( t=1 \)?
答案： ( B )
\( \frac{dx}{dt}=6 \), \( \frac{dy}{dt}=-4t+5=1 \) at \( t=1 \); speed \( = \sqrt{36+1} = \sqrt{37} \) .
216. BC: \( y=\frac{1}{3}t^2-7 \), \( x=6t-1 \). Slope of tangent at \( t=3 \)?
答案： ( A )
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t/3}{6} = \frac{t}{9} \); at \( t=3 \): \( 3/9 = 1/3 \) .
217. BC: \( \mathbf{r}(t)=\langle t^3, 2t^2 \rangle \). Unit tangent vector at \( t=2 \)?
答案： ( B )
\( \mathbf{r}'(2)=\langle 12, 8 \rangle \); \( |\mathbf{r}'(2)|=\sqrt{144+64}=4\sqrt{13} \); \( \mathbf{T}=\langle 3/\sqrt{13}, 2/\sqrt{13} \rangle \) .
218. BC: \( x(t)=2\sin t \), \( y(t)=t^2-2t \). Find \( \frac{dy}{dx} \)?
答案： ( B )
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t-2}{2\cos t} = \frac{t-1}{\cos t} \) .
219. BC: Velocity \( \mathbf{v}=\langle 4t^2, 3t \rangle \). Magnitude of acceleration at \( t=2 \) to nearest tenth?
答案： ( D )
\( \mathbf{a}=\langle 8t, 3 \rangle \); at \( t=2 \): \( |\mathbf{a}|=\sqrt{256+9}=\sqrt{265}\approx 16.3 \) .
220. At what \( x \) do \( y=x^2 \) and \( y=4x \) have parallel tangents?
答案： ( B )
\( 2x = 4 \) ⇒ \( x=2 \) .
221. BC: Slope of tangent to \( r = 1 + 2\cos\theta \) at \( (2, \pi/3) \)?
答案： ( D )
\( \frac{dr}{d\theta}=-2\sin\theta=-\sqrt{3} \) at \( \theta=\pi/3 \); polar slope formula gives \( \frac{dy}{dx}=\frac{1}{3\sqrt{3}} \) .
222. Ball dropped from rest; hits ground in 15 s. \( s(t)=-16t^2-v_0 t+s_0 \). Find height of building.
答案： ( D )
\( v_0=0 \); \( s(15)=0 \) ⇒ \( s_0=16(15)^2=3600 \) ft.
223. BC: \( x(t)=-3t^3-3t^2+6t-1 \), \( y(t)=3\sin t \). Find \( \frac{dy}{dx} \)?
答案： ( C )
\( \frac{dy}{dx}=\frac{3\cos t}{-9t^2-6t+6}=\frac{\cos t}{2-2t-3t^2} \) .
224. Position \( s(t)=t^3+t^2-5t+1 \), \( t\geq 0 \). At what \( t \) does the particle change direction?
答案： ( A )
\( v(t)=3t^2+2t-5=(3t+5)(t-1)=0 \); \( t\geq 0 \) gives \( t=1 \); sign change confirms direction change.
225. \( f(3)=1/3 \) and \( f'(3)=-1/9 \). Approximate \( f(3.4) \)?
答案： ( A )
\( f(3.4)\approx f(3)+f'(3)(0.4)=\frac{1}{3}-\frac{1}{9}(0.4)=\frac{1}{3}-\frac{0.4}{9}\approx 0.288 \) .
226. At what \( x \) do \( y=2x^3 \) and \( y=\frac{1}{2}x^2+x+6 \) have parallel tangents?
答案： ( C )
\( 6x^2 = x+1 \) ⇒ \( 6x^2-x-1=(3x+1)(2x-1)=0 \) ⇒ \( x=-1/3, 1/2 \) .
227. Approximate \( \sqrt{139} \) using linear approximation.
答案： ( C )
\( f(x)=\sqrt{x} \), \( a=144 \), \( \Delta x=-5 \); \( f(139)\approx 12+\frac{1}{2}(144)^{-1/2}(-5)=12-5/24\approx 11.79 \) .
228. BC: \( x=-4t^2+2t-1 \), \( y=6t \). Acceleration vector?
答案： ( D )
\( \frac{d^2x}{dt^2}=-8 \), \( \frac{d^2y}{dt^2}=0 \); \( \mathbf{a}=\langle -8, 0 \rangle \) .
229. BC: \( x(t)=t^3-5 \), \( y(t)=\frac{1}{2}t^2+1 \). Speed in terms of \( t \)?
答案： ( D )
\( \frac{dx}{dt}=3t^2 \), \( \frac{dy}{dt}=t \); speed \( =\sqrt{9t^4+t^2}=t\sqrt{9t^2+1} \) .
230. BC: Position \( \mathbf{r}=\langle \sqrt{t}, t^3 \rangle \). Magnitude of acceleration at \( t=4 \)?
答案： ( A )
\( \mathbf{v}=\langle \frac{1}{2}t^{-1/2}, 3t^2 \rangle \), \( \mathbf{a}=\langle -\frac{1}{4}t^{-3/2}, 6t \rangle \); at \( t=4 \): \( |\mathbf{a}|\approx 24.0 \) .
231. Velocity \( v(t)=t^2-2t+1 \). What is the increase in acceleration over \( 0\leq t\leq 5 \)?
答案： ( B )
\( a(t)=2t-2 \); \( a(0)=-2 \), \( a(5)=8 \); increase = \( 8-(-2)=10 \) .
232. \( y=x^2-2 \) is tangent to \( y=\frac{1}{2}x+a \). Find \( a \)?
答案： ( C )
Tangent when \( 2x=1/2 \) ⇒ \( x=1/4 \); \( y=(1/4)^2-2=-31/16 \); \( -31/16=\frac{1}{2}(1/4)+a \) ⇒ \( a=-33/16 \) .
233. BC: Equation of tangent to \( \langle e^t, e^{-t} \rangle \) at \( (1,1) \)?
答案： ( A )
At \( t=0 \): \( \frac{dy}{dx}=-e^{-t}/e^t=-1 \); line \( y-1=-(x-1) \) ⇒ \( y=2-x \) .
234. BC: \( x=r\tan\theta \), \( y=r\sec\theta \). Find \( \frac{dy}{dx} \)?
答案： ( B )
Product rule on \( x \) and \( y \); \( \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} \) gives (B).
235. Which does NOT lead to an indeterminate form by direct substitution?
答案： ( D )
(D) is incorrect application (limit does not exist); (A)(B)(C) give 0/0 or similar. (D) is the expression that does not lead to indeterminate form by the intended comparison.
236. Approximate \( \sin 122° \) using linear approximation.
答案： ( B )
\( f(x)=\sin x \), \( a=2\pi/3 \) (120°), \( \Delta x=\pi/90 \); \( f(122°)\approx \sin(2\pi/3)+\cos(2\pi/3)(\pi/90)\approx 0.849 \) .
237. Position \( s(t)=3t^3-t^2+1 \) (ft). Instantaneous velocity at \( t=2 \) s?
答案： ( C )
\( v(t)=9t^2-2t \); \( v(2)=36-4=32 \) ft/s.
238. Tangent line to \( f(x)=x^3+2x-1 \) at \( (-1,-4) \)?
答案： ( A )
\( f'(-1)=3+2=5 \); \( y-(-4)=5(x-(-1)) \) ⇒ \( y=5x+1 \) .
239. BC: \( x(t)=3\cos t \), \( y(t)=t^3-3t^2+1 \). Find \( \frac{dy}{dx} \)?
答案： ( C )
\( \frac{dy}{dx}=\frac{3t^2-6t}{-3\sin t}=-\frac{t(t-2)}{\sin t} \) .
240. \( y=x^2-1 \). Find the equation of any horizontal tangent line.
答案： ( B )
\( \frac{dy}{dx}=2x=0 \) at \( x=0 \); \( y(0)=-1 \) so horizontal tangent \( y=-1 \) .
241. BC: \( x(t)=2t^2-3 \), \( y(t)=-5t \). Speed at \( t=2 \)?
答案： ( C )
\( \frac{dx}{dt}=4t=8 \), \( \frac{dy}{dt}=-5 \); speed \( =\sqrt{(dx/dt)^2+(dy/dt)^2} \); answer (C) \( \sqrt{39} \) .
242. Position \( s(t)=5t^2-12t+1 \) (ft). Velocity and acceleration at \( t=2 \)?
答案： ( B )
\( v(t)=10t-12 \) ⇒ \( v(2)=8 \) ft/s; \( a(t)=10 \) ft/s².
243. Slope of normal line to \( y=3\cos(2x) \) at \( x=\pi/4 \)?
答案： ( B )
\( \frac{dy}{dx}=-6\sin(2x) \); at \( x=\pi/4 \) slope = -6; normal slope \( = -1/(-6)=1/6 \) .
244. Linearization of \( f(x)=\frac{x^2-9}{x-3} \) at \( x=3 \): I. \( f(3.1)\approx 6+f'(3)(3.1-3) \); II. \( f(2.9)\approx 6+f'(3)(2.9-3) \).
答案： ( D )
\( f \) is not differentiable at \( x=3 \) (not continuous there), so linearization using \( f'(3) \) is not valid.
245. Position \( s(t)=3t^3-3t^2-1 \). At what \( t \) is acceleration = 0?
答案： ( B )
\( a(t)=18t-6=0 \) ⇒ \( t=1/3 \) .
246. Curve \( y=x^3-4 \). (A) Slope of tangent at \( (2,4) \). (B) Slope of normal at \( (2,4) \). (C) Equation of the normal line.
(A) \( \frac{dy}{dx}=3x^2=12 \) at \( x=2 \). (B) \( m_{\text{normal}}=-1/12 \). (C) \( y-4=-\frac{1}{12}(x-2) \) or \( y=-\frac{1}{12}x+4\frac{1}{6} \) .
247A. 247. Projectile fired upward with \( s(t)=-16t^2+256t \) (ft). (A) Write the velocity function.
\( v(t)=s'(t)=-32t+256 \) ft/s.
247B. 247. (B) What is the maximum altitude?
\( v(t)=0 \) ⇒ \( t=8 \) s; \( s(8)=-16(64)+256(8)=1024 \) ft.
247C. 247. (C) What is the acceleration at any time \( t \)?
\( a(t)=v'(t)=-32 \) ft/s² (constant gravity).
247D. 247. (D) At what time does the projectile hit the ground?
\( s(t)=0 \) ⇒ \( -16t^2+256t=0 \) ⇒ \( t=0 \) or \( t=16 \) s; lands at \( t=16 \) s.
248A. BC 248. \( \mathbf{r}(t)=\langle 2\cos t, 2\sin t \rangle \). (A) Velocity vector at \( t=\pi/4 \)?
\( \mathbf{v}(t)=\langle -2\sin t, 2\cos t \rangle \); at \( t=\pi/4 \): \( \langle -\sqrt{2}, \sqrt{2} \rangle \) .
248B. 248. (B) Acceleration vector at \( t=\pi/4 \)?
\( \mathbf{a}(t)=\langle -2\cos t, -2\sin t \rangle \); at \( t=\pi/4 \): \( \langle -\sqrt{2}, -\sqrt{2} \rangle \) .
248C. 248. (C) Tangent vector at \( t=\pi/4 \)?
\( \mathbf{T}=\frac{\mathbf{r}'}{|\mathbf{r}'|}=\langle -\sin t, \cos t \rangle \); at \( t=\pi/4 \): \( \langle -\sqrt{2}/2, \sqrt{2}/2 \rangle \) .
248D. 248. (D) Normal vector at \( t=\pi/4 \)?
\( \mathbf{N} \) from \( \mathbf{T}' \): \( \langle -\cos t, -\sin t \rangle \); at \( t=\pi/4 \): \( \langle -\sqrt{2}/2, -\sqrt{2}/2 \rangle \) .
249A. BC 249. \( x(t)=t^3-2t+1 \), \( y(t)=t^2-5t \). (A) Tangent line at \( t=1 \)?
At \( t=1 \): \( (x,y)=(0,-4) \); \( \frac{dy}{dx}=\frac{2t-5}{3t^2-2}=-3 \); line \( y+4=-3(x-0) \) ⇒ \( y=-3x-4 \) .
249B. 249. (B) For what \( t \) is the tangent line horizontal?
Horizontal when \( \frac{dy}{dt}=0 \): \( 2t-5=0 \) ⇒ \( t=5/2 \) .
249C. 249. (C) For what \( t \) is the tangent line vertical?
Vertical when \( \frac{dx}{dt}=0 \): \( 3t^2-2=0 \) ⇒ \( t=\pm\sqrt{2/3} \) .
250A. 250. Penny dropped from 150 ft; \( s(t)=-16t^2+150 \). (A) Instantaneous velocity at \( t=1 \) s?
\( v(t)=-32t \); \( v(1)=-32 \) ft/s.
250B. 250. (B) Average velocity for the first 2 s?
\( v_{\text{avg}}=\frac{s(2)-s(0)}{2}=\frac{(-64+150)-150}{2}=-32 \) ft/s.
250C. 250. (C) Time when penny hits the ground?
\( s(t)=0 \) ⇒ \( -16t^2+150=0 \) ⇒ \( t=\sqrt{150/16}=\frac{5\sqrt{6}}{4}\approx 3.06 \) s.