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Application Of Derivatives

Question
CBSEENMA12035127
Wired Faculty App

Find the equation of the tangent and the normal to the curve straight x to the power of 2 over 3 end exponent plus straight y to the power of 2 over 3 end exponent space equals space 2 space space space at space space left parenthesis 1 comma space 1 right parenthesis.

Solution

The equations of curve is
                 straight x to the power of 2 over 3 end exponent plus straight y to the power of 2 over 3 end exponent space equals space 2
Differentiating both sides w.r.t. x, we get,    
2 over 3 straight x to the power of minus to the power of 1 third end exponent end exponent space plus space 2 over 3 straight y to the power of negative 1 third end exponent dy over dx space equals space 0 space space space space space space space space or space space space space space space straight y to the power of negative 1 third end exponent dy over dx space equals negative straight x to the power of negative 1 third end exponent therefore space space space space space space space dy over dx space equals space minus straight x to the power of negative begin display style 1 third end style end exponent over straight y to the power of negative begin display style 1 third end style end exponent space equals space minus straight y to the power of begin display style 1 third end style end exponent over straight x to the power of begin display style 1 third end style end exponent space equals space minus open parentheses straight y over straight x close parentheses to the power of 1 third end exponent At space left parenthesis 1 comma space 1 right parenthesis comma space slope space of space tangent space equals space minus open parentheses 1 over 1 close parentheses to the power of 1 third end exponent space equals space minus 1 therefore space space space space space space space space equation space of space tangent space at space left parenthesis 1 comma space 1 right parenthesis space is space space space space space space space space space space space space space space straight y minus 1 space equals space minus left parenthesis straight x minus 1 right parenthesis comma space space space or space space space straight y minus 1 space equals space minus straight x plus 1 space space space or space space space space straight x plus straight y minus 2 space equals space 0 At space left parenthesis 1 comma space 1 right parenthesis comma space slope space of space normal space space equals space minus fraction numerator 1 over denominator negative 1 end fraction space equals space 1 therefore space space space space space space space equation space of space normal space at space left parenthesis 1 comma space 1 right parenthesis space is space space space space space space space space space space space space space space space space space space space space space space space space space straight y minus 1 space equals space 1 left parenthesis straight x minus 1 right parenthesis space space space space space or space space space space space space space space straight y minus 1 space equals space straight x minus 1 space space space space space space or space space space straight x minus straight y 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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