Question

Let $x=f^{\prime \prime }\left(t\right)\cos t+f^{\prime }\left(t\right)\sin t$ and $y=-f^{\prime \prime }\left(t\right)\sin t+f^{\prime }\left(t\right)\cos t$, then $\int {[{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2]}^{1/2}dt$ equals

• A. $f^{\prime }\left(t\right)+f^{\prime \prime }\left(t\right)+c$
• B. $f^{\prime \prime }\left(t\right)+f^{\prime \prime \prime }\left(t\right)+c$
• C. $f\left(t\right)+f^{\prime \prime }\left(t\right)+c$
• D. $f^{\prime }\left(t\right)-f^{\prime \prime }\left(t\right)+c$

Answer 1

The correct option is C $f\left(t\right)+f^{\prime \prime }\left(t\right)+c$

$\frac{dx}{dt}=f^{\prime \prime \prime }\left(t\right)\cos t-f^{\prime \prime }\left(t\right)\sin t+f^{\prime \prime }\left(t\right)\sin t+f^{\prime }\left(t\right)\cos t$
$\frac{dx}{dt}=\left[f^{\prime \prime \prime }\right(t)+f^{\prime }(t\left)\right]\cos t$

$\frac{dy}{dt}=-f^{\prime \prime \prime }\left(t\right)\sin t-f^{\prime \prime }\left(t\right)\cos t+f^{\prime \prime }\left(t\right)\cos t-f^{\prime }\left(t\right)\sin t$
$\frac{dy}{dt}=-\left[f^{\prime \prime \prime }\right(t)+f^{\prime }(t\left)\right]\sin t$

$\therefore {\left[\begin{array}{c}{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2\end{array}\right]}^{1/2}$
$=\left[\right(f^{\prime \prime \prime }\left(t\right)+f^{\prime }\left(t\right){)}^2({\cos }^{2}t+{\sin }^{2}t){]}^{1/2}$
$=f^{\prime \prime \prime }\left(t\right)+f^{\prime }\left(t\right)$
$\therefore \int {\left[\begin{array}{c}{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2\end{array}\right]}^{1/2}dt=f^{\prime \prime }\left(t\right)+f\left(t\right)+c$