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Application Of Derivatives

Question
CBSEENMA12035237
Wired Faculty App

The slope of the tangent to the curve x = t2 + 3t – 8 , y = 2t2 – 2t – 5 at the point (2, – 1) is
  • 22 over 7
  • 6 over 7
  • 7 over 6
  • fraction numerator negative 6 over denominator 7 end fraction

Solution

B.

6 over 7

The equation of curve is
                                     straight y squared space equals space 4 straight x                                                  ...(1)
The equation of line is
                              straight y equals mx plus 1                                                     ...(2)
Differentiation (1), w.r.t.x, we get,
                                 2 straight y dy over dx space equals space 4 space space space space space or space space space dy over dx space equals space 2 over straight y comma space space space which space is space slope space of space tangent. space
From (2), slope of tangent  = m
therefore space space space space space space space space space space space space space 2 over straight y space equals space straight m space space space space space rightwards double arrow space space space space space space straight y space equals space 2 over straight m therefore space space space space space space from space left parenthesis 1 right parenthesis comma space space space 4 over straight m squared space equals space 4 straight x space space space space rightwards double arrow space space space space space straight x space equals space 1 over straight m squared therefore space space space space space space line space left parenthesis 2 right parenthesis space touches space curve space left parenthesis 1 right parenthesis space at space open parentheses 1 over straight m squared comma space space 2 over straight m squared close parentheses
This point open parentheses 1 over straight m squared comma space space 2 over straight m close parentheses space lies space on space line space left parenthesis 2 right parenthesis
therefore space space space space space space space space space space space space 2 over straight m space equals space straight m space cross times space 1 over straight m squared plus 1 space space space space or space space space space 2 over straight m space equals space 1 over straight m plus 1 therefore space space space space space space space space space 1 over straight m space equals space 1 space space space space space space rightwards double arrow space space space space space space straight m space equals space 1 therefore space space space space space space left parenthesis straight A right parenthesis space is space correct space answer.

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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