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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The two equal sides of an isosceles triangle with fixed base \(b\) are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?

$\begin{array}{1 1} \sqrt {3b}cm^2/sec \\ \sqrt {-3b}cm^2/sec \\ \sqrt {3a}cm^2/sec \\ \sqrt {5b}cm^2/sec \end{array} $

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Toolbox:
  • Area of triangle $=\large\frac{1}{2}$$\times l\times h$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Let $x$ be the equal side of isosceles triangle with fixed base $b$
$AB=AC=x$
$BC=b$
In right $\Delta ABL$,
$AL=\sqrt{AB^2-BL^2}$
$AL=\sqrt{x^2-\large\frac{b^2}{4}}$
Area of $\Delta ABC=A=\large\frac{1}{2}$$\times BC\times AL$
$A=\large\frac{1}{2}$$\times b\times \sqrt{x^2-\large\frac{b^2}{4}}$
Step 2:
It is given that $\large\frac{dx}{dt}$$=3cm/sec$
Now $\large\frac{dA}{dt}=\large\frac{b}{2}\times\frac{1}{2}\times \frac{1}{\sqrt{x^2-b^2/4}}\times 2x\times \large\frac{dx}{dt}$
$\Rightarrow \large\frac{dA}{dt}=\frac{bx}{2}\times \frac{2}{\sqrt{4x^2-b^2}}\times \large\frac{dx}{dt}$
$\large\frac{dA}{dt}=\frac{bx}{2}\times\frac{2}{\sqrt{4x^2-b^2}}\times \frac{dx}{dt}$
Step 3:
When $x=b$ we get,
$\large\frac{dA}{dt}=\frac{b\times b\times 2\times 3}{2\sqrt{4b^2-b^2}}$
$\qquad=\large\frac{3b^2}{\sqrt{3b}}$
$\qquad=\sqrt{3b}$
Hence the area is decreasing at $\sqrt {3b}cm^2/sec$
answered Aug 9, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

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