Question

Calculate dy/dx when x=at^3 and y=bt^2

Answer 1

Heya .

x = at³
so dx/dt = d(at³)/dt
=> 3at². { using d xⁿ/dt = nxⁿ-¹

now y = bt²
dy/dt = d(bt²)/dt
=> 2bt { using d xⁿ/dt = nxⁿ-¹

now dy/dx = dy/dt × dt/dx
so plug the values of dy/dt and dx/dt
we get
dy/dx = 2bt /3at²=> 2b/3at.

Answer 2

x = at^3
dx/dt = 3at^2 ———-(1)

y = bt^2
dy/dt = 2bt ———–(2)

Equation (2) / Equation (1)

dy/dx = 2bt / (3at^2)
dy/dx = 2b / (3at)