KCET · Maths · Application of Derivatives
An enemy fighter jet is flying along the curve, given by \(y=x^2+2\). A soldier is placed at \((3,2)\) wants to shoot down the jet when it is nearest to him. Then, the nearest distance is
- A \(\sqrt{6}\) units
- B 2 units
- C \(\sqrt{5}\) units
- D \(\sqrt{3}\) units
Answer & Solution
Correct Answer
(C) \(\sqrt{5}\) units
Step-by-step Solution
Detailed explanation
Let \(P(x, y)\) be the position of the jet and the
solider is placed at \(A(3,2)\).
\(\Rightarrow \quad A P=\sqrt{(x-3)^2+(y-2)^2}\)
As, \(\quad y=\left(x^2+2\right) \Rightarrow x^2=y-2\)
\(\Rightarrow(A P)^2=(x-3)^2+x^4=z\)
(say)
\(\Rightarrow \quad \frac{d z}{d x}=2(x-3)+4 x^3\) and \(\frac{d^2 z}{d x^2}=2+12 x^2\)
For local points of maxima and minima
\(\begin{aligned} & \frac{d z}{d x}=0 \Rightarrow 2(x-3)+4 x^3=0 \\ & x=1 \text { and } \frac{d^2 z}{d x^2}(\text { at } x=1)=14>0\end{aligned}\)
\(\therefore z\) is minimum, whe \(x=1, y=1+2=3\)
Also, minimum distance
\(=\sqrt{(1-3)^2+(3-2)^2}=\sqrt{5}\) units
solider is placed at \(A(3,2)\).
\(\Rightarrow \quad A P=\sqrt{(x-3)^2+(y-2)^2}\)
As, \(\quad y=\left(x^2+2\right) \Rightarrow x^2=y-2\)
\(\Rightarrow(A P)^2=(x-3)^2+x^4=z\)
(say)
\(\Rightarrow \quad \frac{d z}{d x}=2(x-3)+4 x^3\) and \(\frac{d^2 z}{d x^2}=2+12 x^2\)
For local points of maxima and minima
\(\begin{aligned} & \frac{d z}{d x}=0 \Rightarrow 2(x-3)+4 x^3=0 \\ & x=1 \text { and } \frac{d^2 z}{d x^2}(\text { at } x=1)=14>0\end{aligned}\)
\(\therefore z\) is minimum, whe \(x=1, y=1+2=3\)
Also, minimum distance
\(=\sqrt{(1-3)^2+(3-2)^2}=\sqrt{5}\) units
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