How do you differentiate the following parametric equation: # x(t)=tsqrt(t^2-1), y(t)= t^2-e^(t) #?

Question

How do you differentiate the following parametric equation: #   x(t)=tsqrt(t^2-1),  y(t)= t^2-e^(t)  #?

Emma Aldridge · Accepted Answer

#((2t - e^t)(t^2 - 1)^(1/2))/(2t^2 - 1)#

For the parametric function the derivative is given by

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

rewriting x(t) as # x(t) = t(t^2 - 1)^(1/2) " for ease of differentiating "#

Now have a product of 2 functions, which can be differentiated using the #color(blue)" product rule " #

If f(x) = g(x).h(x) then f'(x) = g(x).h'(x) + h(x).g'(x) #"------------------------------------------------------------"# so x'(t) = t . #d/dt(t^2-1)^(1/2) + (t^2-1)^(1/2).d/dt(t)#

# = t. 1/2(t^2-1)^(-1/2).d/dt(t^2-1) + (t^2-1)^(1/2) .1#

# = t. 1/2(t^2-1)^(-1/2). 2t + (t^2-1)^(1/2) #

#= t^2/(t^2-1)^(1/2) + (t^2 -1)^(1/2) #

rewriting as a single fraction.

# (t^2 + t^2 -1)/(t^2 -1)^(1/2) = (2t^2 -1)/(t^2 -1)^(1/2)# #"-----------------------------------------------------------"#

and #y'(t) = 2t - e^t# #"------------------------------------------------"#

#rArr dy/dx = (y'(t))/(x'(t)) = (2t-e^t)/((2t^2 -1)/(t^2 -1)^(1/2)) #

# =( (2t- e^t)(t^2 -1)^(1/2))/(2t^2 - 1) #

William Abdalla · Answer

undefined To differentiate the parametric equations $x(t) = t\sqrt{t^2 - 1}$ and $y(t) = t^2 - e^t$, you can use the chain rule. Differentiate each equation with respect to $t$ separately to find $dx/dt$ and $dy/dt$, which represent the rates of change of $x$ and $y$ with respect to $t$, respectively.