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 sec^{3} θ and y = b tan^{3} 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 sec^{3} 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
Question 1 :
Differentiate the following
x = a cos^{3}t, y = a sin^{3}t
Solution :
x = a cos^{3}t
dx/dt = a(3cos^{2}t)(-sint)
dx/dt = -3a cos^{2}t sint----(1)
y = a sin^{3}t
dy/dt = a(3sin^{2}t)(cost)
= 3a sin^{2}t cost ----(2)
Divide (2) by (1)
dy/dx = (3asin^{2}t cost) / (-3acos^{2}t sint)
= -sint/cost
dy/dx = -tant
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