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Application Of Derivatives

Question

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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?

Solution

Let ABC be an isosceles triangle in which AB = AC = x (say), BC = b.
From A, draw

AL perpendicular BC comma

so that

BL space equals space straight b over 2

In right

angle straight d space space space increment space ALB comma

AL space equals space square root of AB squared minus BL squared end root space equals space square root of straight x squared minus straight b squared over 4 end root

Let

increment

be area of

increment ABC.

therefore space space space space space space space increment space equals space 1 half BC space cross times space space AL space equals space 1 half straight b square root of straight x squared minus straight b squared over 4 end root

therefore space space space space fraction numerator straight d increment over denominator dt end fraction space equals space 1 half straight b. space fraction numerator 2 straight x over denominator 2 square root of straight x squared minus begin display style straight b squared over 4 end style end root end fraction dx over dt space equals space fraction numerator bx over denominator 2 square root of straight x squared minus begin display style straight b squared over 4 end style end root end fraction. space left parenthesis negative 3 right parenthesis space space space space space space space space space space space open square brackets because space space space dx over dt space equals space 3 space left parenthesis given right parenthesis close square brackets

equals space fraction numerator negative 3 space bx over denominator 2 square root of straight x squared minus begin display style straight b squared over 4 end style end root end fraction

space When space space space space space straight x space equals space straight b comma space space space fraction numerator straight d increment over denominator dt end fraction space equals space fraction numerator negative 3 straight b space cross times space straight b over denominator 2 square root of straight b squared minus begin display style straight b squared over 4 end style end root end fraction space equals space fraction numerator negative 3 straight b squared over denominator square root of 3 space straight b squared end root end fraction space equals space minus square root of 3 space straight b squared end root space equals space minus square root of 3 space straight b therefore space space space space required space rate space of space decrease space space equals space square root of 3 space straight b space cm squared divided by sec.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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