Solve the problem.
-Find the point on the curve $x = 3 \sin t , y = \cos t , - \frac { \pi } { 2 } \leq t \leq \frac { \pi } { 2 }$, closest to the point $\left( \frac { 4 } { 3 } , 0 \right)$. (Hint: Minimize the square of the distance as a function of $t$.)
A) $\left( \frac { 3 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right)$
B) $( - 3,0 )$
C) $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 3 } { 2 } \right)$
D) $( 3,0 )$

Question

Quizplus · Accepted Answer

The distance squared between a point on the curve and the point $\left( \frac { 4 } { 3 } , 0 \right)$ is given by $d^2 = \left( x - \frac { 4 } { 3 } \right)^2 + y^2$. Substituting $x = 3 \sin t$ and $y = \cos t$, we get $d^2 = \left( 3 \sin t - \frac { 4 } { 3 } \right)^2 + \cos^2 t$. To find the minimum distance, we differentiate $d^2$ with respect to $t$ and set the derivative equal to zero. Solving this will give the value of $t$ that minimizes the distance. The point corresponding to this value of $t$ is $\left( \frac { 3 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right)$, which is choice A.

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