A helicopter is flying along the curve given by , (x 0). A solider positioned at the point wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is

Engineering

Mathematics

Tangent and Normal

Question

A helicopter is flying along the curve given by $y - x^{\frac{3}{2}} = 7$ , (x $\geq$ 0). A solider positioned at the point $(\frac{1}{2}, 7)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is

$\frac{1}{2}$

$\frac{1}{3} \sqrt{\frac{7}{3}}$

$\frac{1}{6} \sqrt{\frac{7}{3}}$

$\frac{\sqrt{5}}{6}$

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y – x3/2 = 7(x ≥ 0)

$\frac{dy}{dx} = \frac{3}{2} x^{1 / 2}$

$(\frac{3}{2} \sqrt{x}) (\frac{7 - y}{\frac{1}{2} - x}) = - 1$

$(\frac{3}{2} \sqrt{x}) (\frac{- x^{3 / 2}}{\frac{1}{2} - x}) = - 1$

$\frac{3}{2} \cdot x^{2} = \frac{1}{2} - x$

3x² = 1 – 2x

3x² + 2x – 1 = 0

3x² + 3x – x – 1 = 0

(x + 1)(3x – 1) = 0

∴ x = – 1(rejected)

$x = \frac{1}{3}$

$y = 7 + x^{3 / 2} = 7 + {(\frac{1}{3})}^{3 / 2}$

$ℓ_{AB} = \sqrt{{(\frac{1}{2} - \frac{1}{3})}^{2} + {(\frac{1}{3})}^{3}} = \sqrt{\frac{1}{36} + \frac{1}{27}}$

$= \sqrt{\frac{3 + 4}{9 \times 12}}$

$= \sqrt{\frac{7}{108}} = \frac{1}{6} \sqrt{\frac{7}{3}}$