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MATHMAA

# 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=$$\frac{\pi}{6}$$