이동 거리와 가속도

좌표평면 위를 움직이는 점 $\mathrm{P}$의 시각 $t ( t \geq 1 )$에서의 위치 $( x , y )$가 $x = 2 \ln t , y = f ( t )$로 주어졌으므로 시각 $t$에서의 속도는 $\displaystyle \vec{v} = \left( \frac{2}{t} , f' ( t ) \right)$이고 시각 $t$에서의 가속도는 $\displaystyle \vec{a} = \left( - \frac{2}{t ^{2}} , f'' ( t ) \right)$이다. $t = 2$일 때 점 $\mathrm{P}$의 속도가 $\displaystyle \left( 1 , \frac{3}{4} \right)$이므로 $\displaystyle f' ( 2 ) = \frac{3}{4}$이다. $t = 2$일 때 점 $\mathrm{P}$의 가속도가 $\displaystyle \left( - \frac{1}{2} , a \right)$이므로 $f'' ( 2 ) = a$이다. 점 $\mathrm{P}$가 점 $( 0 , f ( 1 ) )$로부터 움직인 거리가 $s$가 될 때, 시각 $t$는 $\displaystyle t = \frac{s + \sqrt{s ^{2} + 4}}{2}$이고, 주어진 식을 정리하면 $\displaystyle 2 t = s + \sqrt{s ^{2} + 4}$ $( 2 t - s ) ^{2} = s ^{2} + 4$ $4 t ^{2} - 4 ts + s ^{2} = s ^{2} + 4$ $ts = t ^{2} - 1$ $\displaystyle \therefore s = t - \frac{1}{t}$$\cdots \text{㉠}$이다. $\displaystyle s = \int _{1} ^{t} | \vec{v} | dt = \int _{1} ^{t} \sqrt{{\left( \frac{2}{t} \right)} ^{2} + \left( f' ( t ) \right) ^{2}} dt$ 에서 양변을 $t$로 미분하면 $\displaystyle \frac{ds}{dt} = \sqrt{\frac{4}{t ^{2}} + ( f' ( t ) ) ^{2}}$이다. 한편 $\text{㉠}$에서 $\displaystyle \frac{ds}{dt} = 1 + \frac{1}{t ^{2}}$ 이므로 $\displaystyle 1 + \frac{1}{t ^{2}} = \sqrt{\frac{4}{t ^{2}} + ( f' ( t ) )} ^{2}$임을 알 수 있고 양변을 제곱하여 정리하면 $\displaystyle 1 + \frac{2}{t ^{2}} + \frac{1}{t ^{4}} = \frac{4}{t ^{2}} + ( f' ( t ) ) ^{2}$ $\displaystyle ( f' ( t ) ) ^{2} = 1 - \frac{2}{t ^{2}} + \frac{1}{t ^{4}} = \left( 1 - \frac{1}{t ^{2}} \right) ^{2}$ $\displaystyle f' ( t ) = 1 - \frac{1}{t ^{2}}$ $\displaystyle \left( \because f' ( 2 ) = \frac{3}{4} \right)$ $\cdots \text{㉡}$ $\text{㉡}$의 식을 $t$로 미분하면 $\displaystyle f'' ( t ) = \frac{2}{t ^{3}}$이고 $\displaystyle a = f'' ( 2 ) = \frac{1}{4}$이다. 따라서 $\displaystyle 60 a = 60 \times \frac{1}{4} = 15$이다. [다른풀이] $x = 2 \ln t , y = f \left( t \right)$ 이므로 $\displaystyle \frac{dx}{dt} = \frac{2}{t} , \frac{dy}{dt} = f' \left( t \right)$ $\displaystyle \frac{d ^{2} x}{dt ^{2}} = - \frac{2}{t ^{2}} , \frac{d ^{2} y}{dt ^{2}} = f'' \left( t \right)$ $\displaystyle {\vec{v}} = \left( \frac{2}{t} , f' \left( t \right) \right)$ 에 $t = 2$ 를 대입하면 $\displaystyle {\vec{v}} = \left( 1 , f' \left( 2 \right) \right) = \left( 1 , \frac{3}{4} \right)$이므로 ∴ $\displaystyle f' \left( 2 \right) = \frac{3}{4}$ $\displaystyle {\vec{a}} = \left( - \frac{2}{t ^{2}} , f'' \left( t \right) \right)$ 에 $t = 2$를 대입하면 $\displaystyle {\vec{a}} = \left( - \frac{1}{2} , f'' \left( 2 \right) \right) = \left( - \frac{1}{2} , a \right)$ ∴ $a = f'' \left( 2 \right)$ $\displaystyle s = \int _{1} ^{\frac{s + \sqrt{s ^{2} + 4}}{2}} {\sqrt{\left( \frac{2}{t} \right) ^{2} + \left\{ f' \left( t \right) \right\} ^{2}}} dt$ 에서 $\displaystyle g \left( s \right) = \frac{s + \sqrt{s ^{2} + 4}}{2}$라 하자 $\displaystyle s = \int _{1} ^{g \left( s \right)} {\sqrt{\left( \frac{2}{t} \right) ^{2} + \left\{ f' \left( t \right) \right\} ^{2}}} dt$ 양변을 $s$에 대하여 미분하면 $\displaystyle 1 = \sqrt{\left( \frac{2}{g \left( s \right)} \right) ^{2} + \left\{ f' \left( g \left( s \right) \right) \right\} ^{2}} \cdot \left( g' \left( s \right) \right)$ 양변을 제곱하면 $\displaystyle \left[ \left( \frac{2}{g \left( s \right)} \right) ^{2} + \left\{ f' \left( g \left( s \right) \right) \right\} ^{2} \right] \cdot \left( g' \left( s \right) \right) ^{2} = 1$ $\cdots$$\cdots$㉠ 위의 식을 미분하면 $\displaystyle \left[ 2 \left( \frac{2}{g \left( s \right)} \right) \cdot \left( - \frac{2}{\left\{ g \left( s \right) \right\} ^{2}} \right) \cdot g' \left( s \right) + 2 \left\{ f' \left( g \left( s \right) \right) \right\} \cdot f'' \left( g \left( s \right) \right) \cdot g' \left( s \right) \right] \cdot \left( g' \left( s \right) \right) ^{2}$ $\displaystyle + \left[ \left( \frac{2}{g \left( s \right)} \right) ^{2} + \left\{ f' \left( g \left( s \right) \right) \right\} ^{2} \right] \cdot 2 g' \left( s \right) \cdot g'' \left( s \right) = 0$ $\cdots$$\cdots$㉡ 한편, $\displaystyle g' \left( s \right) = \frac{1}{2} \cdot \left( 1 + \frac{2 s}{2 \sqrt{s ^{2} + 4}} \right) = \frac{1}{2} + \frac{s}{2 \sqrt{s ^{2} + 4}}$ $\displaystyle g'' \left( s \right) = \frac{1}{2} \cdot \frac{\sqrt{s ^{2} + 4} - s \cdot \frac{2 s}{2 \sqrt{s ^{2} + 4}}}{s ^{2} + 4} = \frac{2}{\left( s ^{2} + 4 \right) \sqrt{s ^{2} + 4}}$ $\displaystyle g \left( s \right) = \frac{s + \sqrt{s ^{2} + 4}}{2} = 2$ 를 정리하면 $\displaystyle s = \frac{3}{2}$ $\displaystyle g \left( \frac{3}{2} \right) = 2$ , $\displaystyle g' \left( \frac{3}{2} \right) = \frac{4}{5}$ , $\displaystyle g'' \left( \frac{3}{2} \right) = \frac{16}{125}$ ㉠에 $\displaystyle s = \frac{3}{2}$를 대입하면 $\displaystyle \left[ \left( \frac{2}{g \left( \frac{3}{2} \right)} \right) ^{2} + \left\{ f' \left( g \left( \frac{3}{2} \right) \right) \right\} ^{2} \right] \cdot \left( g' \left( \frac{3}{2} \right) \right) ^{2} = 1$ $\displaystyle \left[ \left( \frac{2}{2} \right) ^{2} + \left\{ f' \left( 2 \right) \right\} ^{2} \right] \cdot \left( \frac{4}{5} \right) ^{2} = 1$ $\displaystyle f' \left( 2 \right) = \frac{3}{4}$ ㉡에 $\displaystyle s = \frac{3}{2}$를 대입하면 $\displaystyle \left[ 2 \left( \frac{2}{g \left( \frac{3}{2} \right)} \right) \cdot \left( - \frac{2}{\left\{ g \left( \frac{3}{2} \right) \right\} ^{2}} \right) \cdot g' \left( \frac{3}{2} \right) \right.$ $\displaystyle \begin{aligned} + 2 \left\{ f' \left( g \left( \frac{3}{2} \right) \right) \right\} \cdot f'' \left( g \left( \frac{3}{2} \right) \right) \cdot g' \left( \frac{3}{2} \right) ] \cdot \left( g' \left( \frac{3}{2} \right) \right) ^{2} \\ + \left[ \left( \frac{2}{g \left( \frac{3}{2} \right)} \right) ^{2} + \left\{ f' \left( g \left( \frac{3}{2} \right) \right) \right\} ^{2} \right] \cdot 2 g' \left( \frac{3}{2} \right) \cdot g'' \left( \frac{3}{2} \right) = 0 \end{aligned}$ $\displaystyle \left[ 2 \left( \frac{2}{2} \right) \cdot \left( - \frac{2}{\left\{ 2 \right\} ^{2}} \right) \cdot \frac{4}{5} + 2 \left\{ f' \left( 2 \right) \right\} \cdot f'' \left( 2 \right) \cdot \frac{4}{5} \right] \cdot \left( \frac{4}{5} \right) ^{2}$ $\displaystyle + \left[ \left( \frac{2}{2} \right) ^{2} + \left\{ f' \left( 2 \right) \right\} ^{2} \right] \cdot 2 \cdot \frac{4}{5} \cdot \frac{16}{125} = 0$ ∴ $\displaystyle a = f'' \left( 2 \right) = \frac{1}{4}$ ∴ $60 a = 15$