Question

The shortest distance between the point(1.5, 0) and the curve $\mathrm y=\sqrt{\mathrm x}$.

Answer 1

Hint: To solve this, we will assume a point on the curve in parametric form and then find its distance from the point(1.5, 0). To find the shortest distance, we will differentiate the distance and equate it to 0, so that we get the minima.

Complete step-by-step answer:
Let y = t, so according to the equation x=$t^2$.
Let a point P($t^2$, t) be a point on the curve. The distance between the point P and (1.5, 0) is-
$\mathrm D=\sqrt{\left({\mathrm x}_2-{\mathrm x}_1\right)^2+\left({\mathrm y}_2-{\mathrm y}_1\right)^2}\\\mathrm D=\sqrt{\left(\mathrm t^2-\dfrac32\right)^2+\left(\mathrm t-0\right)^2}\\\\\\$
When D is minimum, $D^2$ = k is also minimum. So we will differentiate k with respect to t-
$\dfrac{\operatorname d\mathrm k}{\operatorname d\mathrm t}=2\left(\mathrm t^2-\dfrac32\right)\left(2\mathrm t\right)+2\mathrm t=0\\2\mathrm t\left(2\mathrm t^2-\dfrac32+1\right)=0\\2\mathrm t\left(2\mathrm t^2-\dfrac12\right)=0\\\mathrm t=0,\pm\dfrac12$
Substituting t=0 in D-
$\mathrm D=\sqrt{\left(0-\dfrac32\right)^2+0}\\\mathrm D=\dfrac32$
Substituting t=½ in D-
$D=\sqrt{\left(\dfrac12-\dfrac32\right)^2+\left(\dfrac12\right)^2}\\D=\;\sqrt{1+\dfrac14}=\dfrac{\sqrt5}2$
Substituting t= -½ in D-
$\mathrm D=\sqrt{\left(-\dfrac12-\dfrac32\right)^2+\left(-\dfrac12\right)^2}\\\mathrm D=\sqrt{4+\dfrac14}=\sqrt{\dfrac94}=\dfrac32$

Clearly, the shortest distance between the point and the curve is-
$\mathrm D=\dfrac{\sqrt5}2\;\mathrm{units}$
This is the required answer.

Note: Instead of checking every possible value of t to find the shortest distance, one can differentiate k again to find the value of t. Also, one common mistake is that students assume that the shortest distance is at t=0, which is incorrect.