Math, asked by gautamverma760, 7 months ago

10. The equation of the normal to the curve
y=sinx at the point (π,O) is :
O x+y=π
O x+y+π=0
O x-y=π
O x-y+π=0​

Answers

Answered by deepikaarya
1

Answer:

The equation of the normal to the curve y=sinx at the point (π,O) is : O x+y=π

Answered by pulakmath007
35

SOLUTION

TO CHOOSE THE CORRECT OPTION

The equation of the normal to the curve

y = sinx at the point (π,O) is :

  • x + y = π

  • x + y + π = 0

  • x - y = π

  • x - y + π = 0

EVALUATION

Here the given equation of the curve is

 \sf{y =  \sin x \: }

Differentiating both sides with respect to x we get

 \displaystyle \sf{ \frac{dy}{dx} =  \cos x  \: }

 \implies \displaystyle \sf{ \frac{dx}{dy} =  \sec x  \: }

 \therefore \:  \displaystyle \sf{ \frac{dx}{dy} \bigg|_{(\pi, 0)}  =  \sec \pi   =  - 1\: }

Hence the required equation of the normal to the given curve is

 \sf{y - 0 =  -   \displaystyle \sf{ \frac{dx}{dy} \bigg|_{(\pi, 0)}   \bigg(x - \pi \bigg)\: } \: }

 \implies \sf{y - 0 = (x - \pi) \: }

 \implies \sf{ x - y = \pi\: }

FINAL ANSWER

Hence the required equation of the normal is

  \sf{ x - y = \pi\: }

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