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Parametric Curves

Parametric Curves - Tangents

In Parametric Curves , let us discuss slope of tangent to the curve and equations of tangents . Before that let us discuss about chain rule of dy/dx , which is the slope of tangent.

  Suppose x=f(t) and y=g(t) be both differentiable functions , then slope of the parametric curve is

\(\frac{dy}{dx}\)=\(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) ........(1)

The  curve has a horizontal tangent when \(\frac{dy}{dt}\) = 0 , provided that \(\frac{dx}{dt}\neq\)0 and it has vertical tangents when \(\frac{dx}{dt}\)=0 provided that \(\frac{dy}{dt}\neq\)0 .

Sketching Parametric Curves , this information is useful.

To find concave up or concave down we need to find second derivative \(\frac{d^{2}y}{dx^{2}}\).

\(\frac{d^{2}y}{dx^{2}}\) =\(\frac{d}{dx}(\frac{dy}{dx})\)

   =\(\frac{d}{dt}(\frac{dy}{dx})*\frac{dt}{dx}\)

Now let us find few problems on equations of tangent, concave up/down , horizontal and vertical tangent points.

1Q: Find \(\frac{dy}{dx}\) of  the following

  a) x=tsin(t) , y=\(t^{2}\)+t

  b) x= \(\frac{1}{t}\), y=\(\sqrt(t)e^{-t}\)

 Solutions:

 a)  x=tsin(t) , y=\(t^{2}\)+t

   \(\frac{dx}{dt}\)= sin(t) + t cos(t)

 \(\frac{dy}{dt}\) = 2t+1

 \(\frac{dy}{dx}\)=\(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

 
 \(\frac{dy}{dx}\)=\(\frac{2t+1}{sin(t)+t*cos(t)}\)

 b) x= \(\frac{1}{t}\), y=\(\sqrt(t)e^{-t}\)

 \(\frac{dx}{dt}\) =\(\frac{-1}{t^{2}}\)

 \(\frac{dy}{dt}\) = \(\frac{1}{2\sqrt(t)}*e^{-t} + \sqrt(t)*-e^{-t}\)

 = \(\frac{1}{2\sqrt(t)}*e^{-t} - \sqrt(t)e^{-t}\)

 \(\frac{dy}{dx}\) = \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

   = \(\frac{\frac{1}{2\sqrt{t}}*e^{-t}-\sqrt{t}e^{-t}}{\frac{-1}{t^{2}}}\)

 =\(\frac{\frac{e^{-t}*(1-2t)}{2\sqrt{t}}}{\frac{-1}{t^{2}}}\)

 =\(\frac{e^{-t}*(2t-1)*t^{\frac{3}{2}}}{2}\)



2Q : Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter

 a) x=1+4t-\(t^{2}\), y= 2- \(t^{3}\) at t=1

 b) x= t- \(t^{-1}\), y= 1+\(t^{2}\) at t=1

 c) x=t cos(t) , y= t sin(t)  at t=\(\pi\)

 d) x= \(sin^{3}t\), y= \(cos^{3}t\) at t=\(\frac{\pi}{6}\)

 solutions:

 a) x=1+4t-\(t^{2}\), y= 2- \(t^{3}\) at t=1

 \(\frac{dx}{dt}\)= 4-2t

 \(\frac{dy}{dt}\)= -\(3t^{2}\)

\(\frac{dy}{dx}\)=\(\frac{-3t^{2}}{4-2t}\)

 at t= 1 slope of tangent

 
\(\frac{dy}{dx}\) = \(\frac{-3}{2}\)

 t=1 we get x= 1+4-1 =4 , y= 2-1 =1

 at t=1 point (4, 1) slope =\(\frac{-3}{2}\)

 Equation of tangent :

 y - 1 = \(\frac{-3}{2}\)(x - 4)

 y = \(\frac{-3}{2}x\) + 6 +1

 y=\(\frac{-3}{2}x\) + 7

 Therefore equation of tangent is 

    y=\(\frac{-3}{2}x\) + 7

 
b) x = t - \(t^{-1}\), y= 1 + \(t^{2}\) at t=1

 \(\frac{dx}{dt}\) = 1 + \( t^{-2} \)

 \(\frac{dy}{dt}\) = 2t

 \(\frac{dy}{dx}\)= \(\frac{2t}{1 + t^{-2}}\)

 slope of tangent at t = 1

 \(\frac{dy}{dx}\) = \(\frac{2}{2}\) = 1

 t = 1 we get x= 1 - 1 = 0, y= 1 + 1 = 2

 at t=1 we get the point as (0, 2) slope = 1

 Equation of tangent at t=1 is

 y - 2 = 1( x - 0)

 y = x + 2

 Therefore equation of tangent at t=1 for given parametric equations is y = x + 2 .

 c) x=t cos(t) , y= t sin(t)  at t=\(\pi\)

 d) x= \(sin^{3}t\), y= \(cos^{3}t\) at t=\(\frac{\pi}{6}\)


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