Question

What is the shortest distance from the point (-3, 3) to the curve $y={{\left( x-3 \right)}^{3}}$?

Answer 1

Hint: Assume the point (x, y) on the curve which is at the shortest distance from the point (-3, 3). Now, assume this shortest distance as ‘s’ and use the distance formula given as $s=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ and square both sides of the relation. Here $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ denotes the points between which the distance is to be found. Substitute the value of y in terms of x and differentiate both sides of the expression for x. Equate $\dfrac{ds}{dx}=0$ and find the value of x using the scientific calculator. After finding the value of x find the value of y and substitute then in the distance formula again to get the answer.

Complete step-by-step solution:
Here we have been provided with the curve $y={{\left( x-3 \right)}^{3}}$ and we are asked to find the shortest distance of this curve from the point (-3, 3).
Now, let us assume a point (x, y) on the curve which is at the shortest distance from the point (-3, 3). Using the distance formula $s=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$, where $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ denotes the points between which the distance is to be found, we get,
$\begin{align}
  & \Rightarrow s=\sqrt{{{\left( x-\left( -3 \right) \right)}^{2}}+{{\left( y-3 \right)}^{2}}} \\
 & \Rightarrow s=\sqrt{{{\left( x+3 \right)}^{2}}+{{\left( y-3 \right)}^{2}}} \\
\end{align}$
On squaring both the sides we get,
$\Rightarrow {{s}^{2}}={{\left( x+3 \right)}^{2}}+{{\left( y-3 \right)}^{2}}$
Substituting the value of y in terms of x using the equation of curve provided we get,
$\begin{align}
  & \Rightarrow {{s}^{2}}={{\left( x+3 \right)}^{2}}+{{\left( {{\left( x-3 \right)}^{3}}-3 \right)}^{2}} \\
 & \Rightarrow {{s}^{2}}={{\left( x+3 \right)}^{2}}+{{\left( x-3 \right)}^{6}}+9-6{{\left( x-3 \right)}^{3}} \\
\end{align}$
For distance (s) to be minimum, we must have its derivative equal to 0, so differentiating both the sides with respect to x and substituting it equal to 0 we get,
$\begin{align}
  & \Rightarrow \dfrac{d\left( {{s}^{2}} \right)}{dx}=\dfrac{d\left[ {{\left( x+3 \right)}^{2}}+{{\left( x-3 \right)}^{6}}+9-6{{\left( x-3 \right)}^{3}} \right]}{dx}=0 \\
 & \Rightarrow \dfrac{d\left( s \right)}{dx}=\dfrac{1}{2s}\left[ 2\left( x+3 \right)+6{{\left( x-3 \right)}^{5}}-18{{\left( x-3 \right)}^{2}} \right]=0 \\
 & \Rightarrow 2\left( x+3 \right)+6{{\left( x-3 \right)}^{5}}-18{{\left( x-3 \right)}^{2}}=0 \\
\end{align}$
Substituting $\left( x-3 \right)=k$ we get,
$\begin{align}
  & \Rightarrow 2\left( k+3+3 \right)+6{{k}^{5}}-18{{k}^{2}}=0 \\
 & \Rightarrow 6{{k}^{5}}-18{{k}^{2}}+2k+12=0 \\
\end{align}$
Now, the above equation is a degree 5 equation and we do not see any integer value satisfying it using the hit and trial method that means roots are decimal. Therefore, we need to use the scientific calculator to find the root, on doing so we get one real root and four imaginary roots so we need to reject the imaginary solutions and consider the real root. So we get,
$\begin{align}
  & \Rightarrow k\approx -0.7219 \\
 & \Rightarrow x-3\approx -0.7219 \\
 & \Rightarrow x\approx 2.2781 \\
\end{align}$
Substituting the above value of x in equation of the curve to determine y we get,
$\begin{align}
  & \Rightarrow y\approx {{\left( 2.2781-3 \right)}^{3}} \\
 & \Rightarrow y\approx -0.3762 \\
\end{align}$
Finally substituting the above obtained values of x and y in the distance formula we get,
$\begin{align}
  & \Rightarrow s=\sqrt{{{\left( 2.2781+3 \right)}^{2}}+{{\left( -0.3762-3 \right)}^{2}}} \\
 & \Rightarrow s=\sqrt{{{\left( 5.2781 \right)}^{2}}+{{\left( -3.3762 \right)}^{2}}} \\
 & \Rightarrow s\approx 6.2655 \\
\end{align}$
Hence, the shortest distance of the given point from the curve is nearly 6.2655 units.

Note: Note that here we do not have any other option to find the real value of x because it is a degree 5 polynomial and we would have only found it on paper by the hit and trial method if there would have been any small integral value of x. It is difficult to think of the root like 2.2781. In engineering mathematics there are methods like Newton – Rhapson method, Regula – Falsi method which are used to find the roots of these types of equations. However, the use of calculators will also be there due to some difficult decimal calculations.