Equation of the normal to the curve y = sinx at the origin is ???
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The curve is
y = sinx ...(i)
Now, differentiating both sides of (i) with respect to x, we get
dy/dx = cosx
⇒ dx/dy = secx
⇒ - dx/dy = - secx
So, (- dx/dy) at the point (0, 0) is
= - sec0
= - 1
∴ at the origin (0, 0), the required normal be
y - 0 = - 1 (x - 0)
⇒ y = - x
⇒ x + y = 0
Hope it helps! (:
The curve is
y = sinx ...(i)
Now, differentiating both sides of (i) with respect to x, we get
dy/dx = cosx
⇒ dx/dy = secx
⇒ - dx/dy = - secx
So, (- dx/dy) at the point (0, 0) is
= - sec0
= - 1
∴ at the origin (0, 0), the required normal be
y - 0 = - 1 (x - 0)
⇒ y = - x
⇒ x + y = 0
Hope it helps! (:
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