ap-calculus-ab

2001 free-response_formB

6 maths questions

Q1 Chain Rule Asymptote Determination View
$f ( x ) = \frac { 2 x } { \sqrt { x ^ { 2 } + x + 1 } }$
a) $x ^ { 2 } + x + 1 > 0$
$x ^ { 2 } + x + ( 1 / 2 ) ^ { 2 } + 1 > 0$
$( x + 1 / 2 ) ^ { 2 } + 1 - 1 / 4 > 0$
$( x + 1 / 2 ) ^ { 2 } + 3 / 4 > 0$ for oll real $x$
Domain: All real $x$
b) see graph
c) $\lim _ { x \rightarrow \infty } \frac { 2 x } { \sqrt { x ^ { 2 } + x + 1 } } = \lim _ { x \rightarrow \infty } \frac { 2 x } { \sqrt { \frac { x } { x ^ { 2 } } + x + 1 } } = \lim _ { x \rightarrow \infty } \frac { 2 } { \sqrt { 1 + \frac { 1 } { x } + \frac { 1 } { x ^ { 2 } } } } = \frac { 2 } { \sqrt { 1 + 0 + 0 } } = 2$
$\lim _ { x \rightarrow - \infty } \frac { 2 x } { \sqrt { x } ^ { 2 } + x + 1 } = \lim _ { x \rightarrow - \infty } \frac { 2 } { \sqrt { 1 + \frac { 1 } { x } + \frac { 1 } { x ^ { 2 } } } } = - \frac { 2 } { \sqrt { 1 + 0 + 0 } } = - 2$
$y = 2 , y = - 2$
d) $f ^ { \prime } ( x ) = \frac { x + 2 } { \left( x ^ { 2 } + x + 1 \right) ^ { 3 / 2 } }$
$f ^ { \prime } ( x ) = 0$
$x + 2 = 0$
$x = - 2$
$f ( - 2 ) = - \frac { 4 } { \sqrt { 3 } } = - \frac { 4 \sqrt { 3 } } { 3 }$
Range: $\{ 4 : - 4 \sqrt { 3 } / 3 \leq 4 < 2 \}$
$V ( t ) = t \cos t , t \geq 0 , y ( 0 ) = 3$
a) $t \cos t > 0,0 \leq t \leq 5$
$t \cos t = 0$
$t = 0$
$$\begin{gathered} \cos t = 0 \quad t \quad \cdots \quad t \\ t = \frac { \pi } { 2 } , \frac { 3 \pi } { 2 } \quad a _ { - } \frac { \pi } { 2 } \quad \frac { \pi } { 2 } \quad E \end{gathered}$$
$( 0 , \pi / 2 ) , ( 3 \pi / 2,5 )$
b) $$\begin{aligned} \Lambda ( t ) & = V ( t ) \\ \Lambda ( t ) & = \cos t + t - \sin t ) \\ & = \cos t - t \sin t \end{aligned}$$
() $Y ( t ) = \int Y ( t ) d t$
$$\begin{aligned} & y ( t ) = \int t \cos t d t \text { let } \quad \frac { d } { d } = t \quad \frac { d v } { d } \\ & = \cos t \\ & = t \sin t - \int \sin t d t \quad \frac { d t } { d t } \quad v \\ & \end{aligned}$$
$Y ( t ) = + \sin t - ( - \cos t ) + c$
$= t \sin t + \cos t + c$
$y | 0 | = 3$
$j = \cos 0 + c$
$c = 2$
$Y ( t ) = t \sin t + \cos t + 2$
d) $V ( t ) = t \cos t$
$t \cos t = 0$
$t \neq 0 \quad \cos t = 0$
$$t = \frac { \pi } { 2 } , \frac { 3 \pi } { 2 } , \ldots$$
$Y \left( \frac { \pi } { 2 } \right) = \frac { \pi } { 2 } \sin \frac { \pi } { 2 } + \cos \frac { \pi } { 2 } + 2 = \frac { \pi } { 2 } + 0 + 2 = \frac { \pi } { 2 } + 2$
Q3 Implicit equations and differentiation Verify implicit derivative and find tangent line features View
$- 8 x ^ { 2 } + 5 x y + y ^ { 3 } = - 149$
a) $- 16 x + 5 y + 5 x a y + 3 y ^ { 2 } d y = 0$
$\operatorname { dy } \left( 5 x + 5 y ^ { 2 } \right) = 16 x - 5 y$
$d x$
$d u = 16 x - 54$
$d x 5 x + 3 y ^ { 2 }$
b) at (4,-1) $\frac { d y } { d x } = \frac { ( 6 \mid 4 ) - 5 ( - 1 ) } { 5 ( 4 ) + 3 ( 1 ) } = \frac { 64 + 5 } { 23 } = \frac { 69 } { 23 } = 3$
Equation of tangent:
$u - ( - 1 ) = 3 ( x - 4 )$
$4 + 1 = 3 x - 12$
$y = 3 x - 13$
C) $( 4,2 , k )$
$k = 3 ( 4.2 ) - 13$
$k = - 0.4$
d) $- 8 ( 4.2 ) ^ { 2 } + 5 ( 4.2 ) ( k ) + k ^ { 3 } = - 149$
$- 141.12 + 21 k + k 3 = - 149$
$k ^ { 3 } + 21 k + 7.88 = 0$
CJUsing G.D.C.
$k = - 0.373$ to 3.d.p
a) $f ( x ) = a ( x )$
$x ^ { 2 } = 2 ^ { x }$
$x ^ { 2 } - 2 ^ { x } = 0$
Using G. D. C.
(-0,767,0)
$( 2,0 )$
$( 4,0 )$
b) $\int _ { - 0.167 } ^ { 2 } \left( 2 ^ { x } - x ^ { 2 } \right) d x + \int _ { 2 } ^ { 4 } \left( x ^ { 2 } - 2 ^ { x } \right) d x$
C) $R ( x ) = 5 - y _ { 2 } = 5 - x ^ { 2 }$
$r | x | = 5 - 4 i = 5 - 2 ^ { x }$
volume $= \pi \int _ { - 0.767 } ^ { 2 } \left[ \left( 5 - x ^ { 2 } \right) ^ { 2 } - \left( 5 - x ^ { 2 } \right) ^ { 2 } \right] d x$
Q5 Applied differentiation Applied modeling with differentiation View
a) $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$
using similar triangles:
$\vec { V } = \frac { 1 } { 3 } \pi \left( \frac { 1 } { 3 } h \right) ^ { 2 } h = \frac { 1 } { 27 } \pi h ^ { 3 }$
$$\frac { d h } { d t } = h - 12 , V = \frac { 1 } { 3 } \pi r ^ { 2 } h$$
b) we want dv when $h = 3$
$$\begin{aligned} V & = \frac { 1 } { 27 } \pi h ^ { 3 } d t \\ d y & = \frac { 1 } { a } \pi h ^ { 2 } d h \\ d t & = \frac { 1 } { a t } \pi h ^ { 2 } ( h - 12 ) \\ & = \frac { 1 } { 9 } \pi \left( 3 ^ { 2 } \right) ( - 9 ) \\ & = - 9 \pi \end{aligned}$$
$- 9 \pi \mathrm { ft } ^ { 3 } / \min$
c) we want $\frac { d y } { d t }$ when $h = 3$
$$\begin{aligned} & \pi R ^ { 2 } = 400 \pi \\ & R ^ { 2 } = 400 \\ & R = 20 \end{aligned}$$
volume of culinder $= \pi R ^ { 2 } 4$
$\frac { d V } { d t } = 2 \pi R \varphi \frac { d R } { d t } + \pi R ^ { 2 } \frac { d y } { d t }$
$d V = \pi R ^ { 2 } d y \quad d R = 0$ since $R$ is a constant
$d t d t d t$
$9 \pi = 400 \pi \frac { d \| } { d t }$
$d u = q \quad t +$ Imin
d1 400
or:
$$\begin{aligned} & y = 400 \pi y \\ & \frac { d y } { d t } = 400 \pi d y \\ & \frac { d y } { d t } = \frac { 9 \pi } { 400 \pi } = \frac { 9 } { 400 } \end{aligned} \text { Himin }$$
Q6 Stationary points and optimisation Accumulation Function Analysis View
$$h ( x ) = \int _ { 1 } ^ { x } t ( t ) d t$$
a) $h ( 1 ) = \int ^ { \prime } f ( t ) d t = 0$
b) $$\begin{aligned} & h h ^ { \prime } ( x ) = \frac { d } { d x } \int _ { 0 } ^ { x } f ( t ) d t = f ( x ) \\ & h ^ { \prime } ( 4 ) = f ( 4 ) = 2 \end{aligned}$$
oncave up on $( 1,3 )$ and $( 6,7 )$ since $h ^ { \prime } ( x )$ is increa. those intervals
d) h(x) has no relotive minimum on [1,7] since $h ^ { \prime } ( x )$ does
change sign from - ve to tre
$\therefore$ Minimum must accur at either enopoint
$$\begin{aligned} & h ( 1 ) = 0 \\ & h ( 7 ) = \int _ { 1 } ^ { 7 } f ( t ) d t \end{aligned}$$
$h ( 7 ) > h ( 1 ) \quad$ since $h ( x )$ has a relative maximum $a t \quad x = 5$, and connot decrease to zero, $\int ^ { 5 } h ^ { \prime } ( x ) > \int _ { 5 } ^ { 7 } h ^ { \prime } ( x ) d x$