" vil."quad x=sqrt(t),y=t-(1)/(sqrt(t))" at "t=4

The slopee of normal to the curve x=sqrt(t) and y=t-(1)/(sqrt(t)) at t=4 is . . .

Equations of the tangent and normal to the curve x=sqrtt, y=t- (1)/(sqrt t) at the point t=4 are respectively

x=sqrt(sin 2t),y=sqrt(cos 2 t)

Find the equations of tangents and normals to the curve at the point on it: x= sqrt t , y= t - frac{1}{sqrt t} at t=4

Differentiate w.r.t. time. (i) y=t^(2) " " (ii) x=t^(3//2)" " (iii) y=(1)/(sqrt(t)) (iv) x=4t^(3) " " (v) y=2sqrt(t) " " (vi) y=2t^(2)+t-1 (vii) y=3sqrt(t)+(2)/(sqrt(t)) (viii) y=t^(3)sin t " " (ix) x=te^(t) (x) x= sqrt(t)(1-t)

If e^x=(sqrt(1+t)-sqrt(1-t))/(sqrt(1+t)+sqrt(1-t)) and tan (y/2)=sqrt((1-t)/(1+t)) then (dy)/dx at t=1/2 is

e^(x)=(sqrt(1+t)-sqrt(1-t))/(sqrt(1+t)+sqrt(1-t)) andtan (y)/(2)=sqrt((1-t)/(1+t)) then (dy)/(dx) at t=(1)/(2) is (a)-(1)/(2) (b) (1)/(2)(c)0(d) none of these

If x=cos^-1(1/(sqrt(1+t^2))) , y=sin^-1(1/(sqrt(t^2+1))) then dy/dx is independent of t.