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Application Of Derivatives

Question

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Use differentials to approximate:
$square root of 49.5 end root$

Solution

Take

straight y space equals space square root of straight x comma space space space space space straight x space equals space 49 comma space space space space space dx space equals space δx space equals space 0.5 space space space so space tha space straight x plus δx space equals space 49.5

Now,

space space space space straight y plus δy space equals space space square root of straight x plus δx end root space space space space space space space space space space space space space rightwards double arrow space space space δy space equals space square root of straight x plus δx end root space minus space straight y space equals space square root of 49.5 end root space minus space 7

rightwards double arrow

square root of 49.5 end root space equals space space δy space plus space 7

...(1)
Now,

δy

is approximately equal to dy
and

dy space equals space dy over dx dx space equals space fraction numerator 1 over denominator 2 square root of straight x end fraction dx space equals space fraction numerator 1 over denominator 2 square root of 49 end fraction left parenthesis 0.5 right parenthesis space equals space fraction numerator 0.5 over denominator 14 end fraction space equals space 0.0357

therefore space space space space from space left parenthesis 1 right parenthesis comma space space space space square root of 49.5 end root space equals space 0.0357 space plus space space 7 space equals space 7.0357

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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