Find the slopes of the tangent and the normal to the following curves at the indicated points: `y=sqrt(x^3)a tx=4` `y=sqrt(x^3)a tx=9` `y=x^3-xa tx=2` `y=2x^2+3sinxa tx=0` `x=a(theta-sintheta),y=a(1+costheta)a ttheta=-pi/2` `x=acos^3theta,y=asin^3thetaa ttheta=pi/4` `x=a(theta-s intheta),y=a(1-costheta)a ttheta=pi/2` `y=(sin2x+cotx+2)^2a tx=pi/2` `x^2+3y+y^2=5a t(1,1)` `x y=6a t(1,6)`

Explanation

`(i)y=sqrt(x^3)\at x=4`
`(dy)/(dx)|_(x=4) = 3/2 (x^(1/2))`
`=3/2 xx 2`
`=3`(slope of tangent)
slope of normal `=-1/3`
`(iv) y=2x^2+3sinxa tx=0`
`(dy)/(dx)|_(x=0) =4x+3cosx `
`=0+3`
`=3`(slope of tangent)
slope of normal `=-1/3`

`(v) x=a(theta-sin theta),y=a(1+cos theta)a t theta=pi/2`
`(dy)/(dx)_(theta=pi/2)=((dy)//(d theta))/((dx)//(d theta))`
`=(-sin theta)/(1-cos theta)`
`= (-sin (pi/2))/(1-cos (pi/2))`
`=1(slope of tangent)`
slope of normal`=-1`

`(x)x y=6 a t(1,6)`
`x*(dy)/(dx)+y=0`
`(dy)/(dx)_(1,6)=-y/x`
`=-6/1`(slope of tangent)
slope of the normal `=1/6`