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Application Of Derivatives

Question
CBSEENMA12035117
Wired Faculty App

Find the equation of the tangent and normal to the given curves at the points given:
y = x2 at (2, 8)

Solution

The equation of curve is y = x2
therefore space space space space space space dy over dx space equals space 3 straight x squared
At (2, 8),  dy over dx space equals space 3 left parenthesis 2 right parenthesis squared space equals space 12 comma which is the slope of tangent
therefore space space space equation space of space tangent space at space left parenthesis 2 comma space 8 right parenthesis space is space space space space space space space space space space straight y minus 8 space equals space 12 left parenthesis straight x minus 2 right parenthesis comma space space space space or space space space space space straight y minus 8 space equals space 12 straight x minus 24 space space space space or space space space space space 12 straight x minus straight y minus 16 space equals space 0 Slope space of space normal space space equals space minus 1 over 12
The equation of normal at (2, 8) is
                         straight y minus 8 space equals space minus 1 over 12 left parenthesis straight x minus 2 right parenthesis space space or space space space 12 straight y space minus 96 space equals space minus straight x plus 2 space space space or space space space straight x plus 12 straight y minus 98 space equals space 0

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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