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प्रश्न
Find the tangent and normal to the following curves at the given points on the curve
y = x sin x at `(pi/2, pi/2)`
उत्तर
y = x sin x at `(pi/2, pi/2)`
Differentiating w.r.t. ‘x’
`("d"y)/("d"x)` = x cos x + sin x
Slope of the tangent ‘m’ = `(("d"y)/("d"x))_(((pi/2, pi/2)))`
= `pi cos pi/2 + sin pi/2` = 1
Slope of the Normal `- 1/"m"` = – 1
Equation of tangent is
y – y1 = m(x – x1)
⇒ `y - pi/2 = 1(x - pi/2)`
⇒ x – y = 0
Equation of Normal is
y – y1 = `- 1/"m"(x - x_1)`
⇒ `y - pi/2 = -1(x - pi/2)`
⇒ `y - pi/2 = - x + pi/2`
⇒ x + y – π = 0
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