CUET Preparation Today
Find the slope of the tangent or normal to given curves at the indicated point as instructed: y=\( {x }^{ 3} \) at (1,1) : normal |
x+3y-4=0 x+8y-7=0 x-y-4=0 x-y-1=0 |
x+3y-4=0 |
$y=x^3$ $⇒\frac{dy}{dx}=3x^2$ 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) $⇒C'=4$ $∴3y=-x+4$ |
