DIFFERENTIATION OF PARAMETRIC FUNCTION

Differentiation of parametric function is another interesting method in the topic differentiation.

In this method we will have two functions known as x and y. Each function will be defined using another third variable. 

Example 1 :

Find  dy/dx for the below parametric functions.

x = a cos θ and y = b sin θ

Solution :

Here the two variables x and y are defined using third variable θ.

By using the function x we can only find dx/dθ and by using the function y we can only find dy/dθ. By dividing these two derivatives we can find dy/dx.

x = a cos θ                       y = b sin θ

dx/dθ = a (-sin θ)             dy/dθ = b cos θ

dx/dθ = - a sin θ

Now we can find dy/dx

dy/dx = (dy/dθ)/(dx/dθ)

dy/dx = (b cos θ)/(- a sin θ)

dy/dx = -b/a (cos θ/ sin θ)

dy/dx = -(b/a) cot θ

Example 2 :

Find dy/dx by using the parametric functions

x = a sec³ θ  and  y = b tan³ θ

Solution :

Here the two variables x and y are defined using third variable θ.

By using the function x we can only find dx/dθ and by using the function y we can only find dy/dθ.

By dividing these two derivatives we can find dy/dx.

x = a sec³ θ 

dx/dθ  =  3a sec² θ(sec θ tan θ) 

=  3a sec³ θ tan θ

y  =  b tan³ θ

dy/dθ  =  3b tan² θ  sec² θ

Now we have divide those two derivatives to find dy/dx

dy/dx = (dy/dθ)/(dx/dθ)

=  (3b tan² θ  sec² θ)/(3a sec³ θ tan θ)

=  (b/a) (tan θ/sec θ)

=  (b/a) (sin θ/cos θ)/(1/cos θ)

=  (b/a) (sin θ/cos θ) x (cos θ/1)

=  (b/a) sin θ

Example 3 :

Differentiate the following

x = a cos³t, y = a sin³t

Solution :

x = a cos³t

dx/dt  =  a(3cos²t)(-sint) 

dx/dt  =  -3a cos²t sint----(1)

y = a sin³t

dy/dt  =  a(3sin²t)(cost)

  =  3a sin²t cost ----(2)

Divide (2) by (1)

dy/dx  =  (3asin²t cost) / (-3acos²t sint) 

  =  -sint/cost

 dy/dx  =  -tant

Example 4 :

Find dy/dx and d²y/dx²

x = t², y = t² + 6t + 5

Solution :

x = t², y = t² + 6t + 5

Finding the first derivative :

dx/dt = 2t

dy/dt = 2t + 6(1) + 0

= 2t + 6

dy/dx = (dy/dt) / (dx/dt)

= (2t+6)/2t

= 2(t + 3) / 2t

= (t + 3) / t

Finding the second derivative :

dx/dt = 2t

d²x/dt² = 2(1)

= 2

dy/dt = 2t + 6

d²y/dt² = 2(1)

= 2

d²y/dx² = (d²y/dt²) / (d²x/dt²)

= 2/2

= 1

Example 4 :

Find dy/dx and d²y/dx²

x = ln t, y = t² + t

Solution :

x = ln t, y = t² + t

Finding the first derivative :

dx/dt = 1/t

dy/dt = 2t + 1

dy/dx = (dy/dt) / (dx/dt)

= (2t + 1)/(1/t)

= t(2t + 1)

Finding the second derivative :

dx/dt = 1/t = t^-1

d²x/dt² = -1t^-1-1

= -t^-2

= -1/t²

d²y/dt² = 2(1) + 0

= 2

(d²y/dt²) / (d²x/dt²) = 2/(-1/t²)

d²y/dx² = -2t²

Example 5 :

A curve C is defined by the parametric equations 

x = t² + t - 1

y = t³ - t²

(a) Find dy dx in terms of t.

(b) Find an equation of the tangent line to C at the point where t = 2.

Solution :

x = t² + t - 1

y = t³ - t²

dx/dt = 2t + 1

dy/dt = 3t² - 2t

dy/dx = (dy/dt) / (dx/dt)

= (3t² - 2t) / (2t + 1)

To find equation of tangent, we need a point and slope of the tangent at the specific point.

Slope at t = 2

= (3(2)² - 2(2)) / (2(2) + 1)

= (3(4) - 4) / (4 + 1)

= (12 - 4)/5

= 8/5

Equation of the line :

y - y₁ = m(x - x₁)

y - 4 = (8/5) (x - 5)

5(y - 4) = 8(x - 5)

5y - 20 = 8x - 40

8x - 5y - 40 + 20 = 0

8x - 5y - 20 = 0

Example 6 :

A curve C is defined by the parametric equations

x = 2 cos t, y = 3 sin t.

(a) Find dy dx in terms of t.

(b) Find an equation of the tangent line to C at the point where t = π/4

Solution :

x = 2 cos t, y = 3 sin t.

a)

dx/dt = -2 sin t and dy/dt = 3 cost

(dy/dt)/(dx/dt) = 3 cos t / (-2 sin t)

dy/dx = (-3/2)(cos t / sin t)

dy/dx at t = π/4

= (-3/2)(cos π/4 / sin π/4)

= (-3/2) (√2/2) / (√2/2)

= -3/2

b)  Equation of tangent :

(y - y₁) = m(x - x₁)

(y - √2) = (-3/2)(x - √2)

2(y - √2) = -3(x - √2)

2y - 2√2 = -3x + 3√2

3x + 2y - 2√2 - 3√2 = 0

3x + 2y - 5√2 = 0

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