Question

Calculus and Geometry How close does the semicircle \(y=\sqrt{16-x^{2}}\) come to the point \((1, \sqrt{3}) ?\) ?

Short answer

The smallest distance from the semicircle to the point \(\left(1,\sqrt{3}\right)\) can be obtained by solving the steps. The derivative ultimately gives the \(x\) coordinate on the semicircle that is closest to the point. After inputting this value into the original distance formula, the minimum distance between the semicircle and the point is calculated.

Step-by-step solution

Determine the equation for the distance

The distance \(d\) between any point \((x, y)\) on the semicircle and the given point \((1, \sqrt{3})\) can be mathematically represented by the Euclidean distance formula which is:\[d = \sqrt{(x-1)^{2} + (y-\sqrt{3})^{2}}.\]Since \(y=\sqrt{16-x^{2}}\), this equation can be substituted in the formula to get \[d = \sqrt{(x-1)^{2} + (\sqrt{16-x^{2}}-\sqrt{3})^{2}}.\]

Differentiate the equation for the minimum distance

In order to find the minimum distance, differentiate the distance equation with respect to \(x\) and then set the derivative equal to zero to find the \(x\) value that gives the minimum distance. Because working with the square root can be tricky, it's easier to work with \((d^{2})'\) instead of \(d'\):\[(d^{2})' = ((x-1)^{2} + (\sqrt{16-x^{2}}-\sqrt{3})^{2})' = 0\]

Substitute the value of \(x\) into the equation of the line

Once the derivative is set to zero and solved, a value for \(x\) will be obtained. Plug this value back into the distance equation from step 1 to find the minimum distance. The \(x\) value giving the minimum distance when subbed into the distance equation will give the final answer.