An enemy fighter jet is flying along the curv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $P(x, y)$ be the position of the jet and the soldier be at $A(3, 2)$.The distance $AP$ is given by $AP = \sqrt{(x - 3)^2 + (y - 2)^2}$.Since t

Vedclass · Accepted Answer

$\sqrt{5}$ units

Vedclass · Answer

(C) Let $P(x, y)$ be the position of the jet and the soldier be at $A(3, 2)$.The distance $AP$ is given by $AP = \sqrt{(x - 3)^2 + (y - 2)^2}$.Since the jet follows the curve $y = x^2 + 2$,we have $y - 2 = x^2$.Substituting this into the distance formula,let $z = (AP)^2 = (x - 3)^2 + (x^2)^2 = (x - 3)^2 + x^4$.To find the minimum distance,we differentiate $z$ with respect to $x$: $\frac{dz}{dx} = 2(x - 3) + 4x^3$.Setting $\frac{dz}{dx} = 0$,we get $4x^3 + 2x - 6 = 0$,which simplifies to $2x^3 + x - 3 = 0$.By inspection,$x = 1$ is a root since $2(1)^3 + 1 - 3 = 0$.Checking the second derivative: $\frac{d^2z}{dx^2} = 12x^2 + 2$. At $x = 1$,$\frac{d^2z}{dx^2} = 14 > 0$,so $x = 1$ gives a minimum.For $x = 1$,$y = (1)^2 + 2 = 3$.The minimum distance is $\sqrt{(1 - 3)^2 + (3 - 2)^2} = \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}$ units.

An enemy fighter jet is flying along the curve given by $y = x^2 + 2$. $A$ soldier is placed at $(3, 2)$ and wants to shoot down the jet when it is nearest to him. The nearest distance is:

Similar Questions

The minimum value of the function $f(x)=2|x-1|+|x-2|$ is

The maximum value of ${\left( {\frac{1}{x}} \right)^x}$ is

Let $f : R \rightarrow R$ be a function defined by $f(x) = ||x+2|-2|x||$. If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$,then $m+n$ is

Find the minimum value of $64 \sec x + 27 \csc x$ for $0 < x < \frac{\pi}{2}$.

For all real values of $x,$ the minimum value of $\frac{1-x+x^{2}}{1+x+x^{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module