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Find the normal equation of curve y = arctan2x at point (0,0)
y=arctan(2x)
y'=2/(1+(2x)^2)=2/(1+4x^2)
Y '(0) = 2 normal slope - 1 / y' (0)
Normal equation
y-0=[-1/y'(0)](x-0)
y=(-1/2)x
The normal equation of the curve y = 2 + 1 / X at x = 1
First, we get the derivative function y '= LNX
The meaning of derivative function is that the slope at x is 0, which is a straight line parallel to X axis
Its normal is a straight line perpendicular to it, and the equation is x = 1
Given that the normal of the point on the curve y = x ^ 2 + X passes through M (- 2,2), the normal equation is solved
Can you write the process clearly?
X-3y + 8 = 0. Curve y = x & sup2; + X. the derivation is y '= 2x + 1. ∵ 2 = (- 2) & sup2; + (- 2).. point m (- 2,2) is on the curve y = x & sup2; + X. when x = - 2, y' = - 3. The tangent equation of the curve at point m is Y-2 = - 3 (x + 2). That is, 3x + y + 4 = 0. The normal of the curve at point m is a straight line passing through point m (- 2,2) and perpendicular to the tangent, and the equation is Y-2 = (1 / 3) (x + 2). That is, x-3y + 8 = 0
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