Notes On Derivatives of Functions in Parametric Form - CBSE Class 12 Maths
Here two relations x = at2   y = 2at The relation between variables x and y is expressed via third variable t. This third variable is known as the parameter. Expressing a relation between two variables x and y in the form x = f(t) and y = g(t) is said to be its parametric form, with t as the parameter. We have to find the dy/dx then we use the chain rule. dy/dt = dy/dx . dx/dt ⇒ dy/dx = (dy/dt)/(dx/dt)  (provided dx/dt ≠ 0) ∴ dy/dx = g'(t)/f'(t)      (provided f'(t) ≠ 0) dy/dx is expressed in terms of the parameter, and it does not directly involve the main variables x and y. Take x = at2     y = 2at Differentiating x = at2 w.r.t. t , we get dx/dt = d/dt(at2) = at . 2 = 2at  Differentiating y = 2at w.r.t. t, we get dy/dt = d/dt(2at) = 2a ∴ dy/dx = (dy/dt)/(dx/dt) = 2a/2at = 1/t

#### Summary

Here two relations x = at2   y = 2at The relation between variables x and y is expressed via third variable t. This third variable is known as the parameter. Expressing a relation between two variables x and y in the form x = f(t) and y = g(t) is said to be its parametric form, with t as the parameter. We have to find the dy/dx then we use the chain rule. dy/dt = dy/dx . dx/dt ⇒ dy/dx = (dy/dt)/(dx/dt)  (provided dx/dt ≠ 0) ∴ dy/dx = g'(t)/f'(t)      (provided f'(t) ≠ 0) dy/dx is expressed in terms of the parameter, and it does not directly involve the main variables x and y. Take x = at2     y = 2at Differentiating x = at2 w.r.t. t , we get dx/dt = d/dt(at2) = at . 2 = 2at  Differentiating y = 2at w.r.t. t, we get dy/dt = d/dt(2at) = 2a ∴ dy/dx = (dy/dt)/(dx/dt) = 2a/2at = 1/t

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