Find the slope of the tangent or normal to given curves at the indicated point as instructed:
y=\( {x }^{ 3} \) at (1,1) : normal
Answer & explanation
Correct answer: option 1
$y=x^3$
$⇒\frac{dy}{dx}=3x^2$
Slope of normal = $\frac{-1}{dy/dx}=\frac{-1}{3x^2}$
$=\frac{-1}{3}$
$⇒y=mx+C$
$⇒y=\frac{-1}{3}x+C$
$⇒3y=-x+C'$ → satisfied by (1, 1)
$⇒C'=4$
$∴3y=-x+4$