Express \( p(x) = l_1^2 + l_2^2 \) in terms of x.
Solution and Explanation
Given:
- Height of first pole = \(22\) m
- Height of second pole = \(16\) m
- Width of the road = \(20\) m
- Distance of point \(R\) from the smaller pole = \(x\) m
Step 1: Find \(l_1\).
The horizontal distance between the taller pole and point \(R\) is
\[ 20-x. \]
Using Pythagoras theorem,
\[ \begin{aligned} l_1^2& amp;=(20-x)^2+22^2\\ & amp;= (20-x)^2+484 \end{aligned} \] Hence, \[ l_1=\sqrt{(20-x)^2+484}. \]
Step 2: Find \(l_2\).
The horizontal distance from \(R\) to the smaller pole is \(x\).
\[ \begin{aligned} l_2^2& amp;=x^2+16^2\\ & amp;=x^2+256 \end{aligned} \] Hence, \[ l_2=\sqrt{x^2+256}. \]
Therefore,
\[ \boxed{ p(x)=\sqrt{(20-x)^2+484}+\sqrt{x^2+256} } \]
Find \( p'(x) \).
Solution and Explanation
From part (i),
\[ p(x)=\sqrt{(20-x)^2+484}+\sqrt{x^2+256}. \]
Differentiating,
\[ \begin{aligned} p'(x) & amp;=\frac{-2(20-x)} {2\sqrt{(20-x)^2+484}} +\frac{2x} {2\sqrt{x^2+256}}\\ & amp;=\frac{x-20} {\sqrt{(20-x)^2+484}} +\frac{x} {\sqrt{x^2+256}} \end{aligned} \]
Hence,
\[ \boxed{ p'(x)= \frac{x-20}{\sqrt{(20-x)^2+484}} + \frac{x}{\sqrt{x^2+256}} } \]
Using integration, find the area of region OAC, in which children are allowed to play games like cricket, football.
Solution and Explanation
Step 1: Form the function.
\[ \begin{aligned} l_1^2+l_2^2 & amp;=\left((20-x)^2+484\right) +\left(x^2+256\right)\\ & amp;=2x^2-40x+1140 \end{aligned} \]
Step 2: Differentiate.
\[ \frac{d}{dx}(l_1^2+l_2^2)=4x-40. \]
For minimum,
\[ 4x-40=0. \] Therefore, \[ x=10. \]
Step 3: Second derivative test.
\[ \frac{d^2}{dx^2}(l_1^2+l_2^2)=4>0. \] Hence the function is minimum.
Answer:
\[ \boxed{x=10\text{ m}} \]
If the 22 m long pole is also replaced by a 16 m long pole, at
what distance from either pole should the ladders be kept so
that the sum of squares of lengths of ladders needed to reach
the top of the pole is minimum ?
Solution and Explanation
Given:
- Both poles have height \(16\) m.
Step 1: Form the function.
\[ \begin{aligned} l_1^2+l_2^2 & amp;=\left((20-x)^2+16^2\right) +\left(x^2+16^2\right)\\ & amp;=2x^2-40x+912. \end{aligned} \]
Step 2: Differentiate.
\[ \frac{d}{dx}(l_1^2+l_2^2)=4x-40. \] Setting it equal to zero, \[ 4x-40=0. \] Hence, \[ x=10. \]
Step 3: Verify minimum.
\[ \frac{d^2}{dx^2}(l_1^2+l_2^2)=4>0. \] Therefore, the minimum occurs at \[ \boxed{x=10\text{ m}}. \]
Thus, the ladders should be placed 10 m from either pole, i.e., exactly at the midpoint of the road.
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