How do you differentiate the following parametric equation: # x(t)=(t+3t^2)e^t , y(t)=t/e^(3t) #?

Question

1537 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-26T10:44:41

# dy/dx=(1-3t)/((1+7t+3t^2)e^(4t)), t!=(-7+-sqrt37)/2.#

#y(t)=t/e^(3t)=te^(-3t)#

#:. y'(y)=t{d/dte^(-3t)}+e^(-3t)d/dt(t)#

#=t(e^(-3t))d/dt(-3t)+e^(-3t)(1)...[because," the Chain Rule]."#

#:. y'(t)=(1-3t)e^(-3t)..........(1).#

#x(t)=(t+3t^2)e^t#

#:. x"(t)=(t+3t^2)d/dt(e^t)+e^td/dt(t+3t^2)#

#=(t+3t^2)e^t+e^t(1+6t)#

#:. x'(t)=(t+3t^2+1+6t)e^t=(1+7t+3t^2)e^t............(2).#

Now, by the Rule for Parametric Diffn., we have,

#dy/dx=(y'(t))/(x'(t)), x'(t)!=0...........(star).#

Now, #x'(t)=0 rArr (1+7t+3t^2)e^t=0#

# rArr 1+7t+3t^2=0, because, e^t!=0, AA t in RR.#

# rArr t={-7+-sqrt(49-4(1)(3))}/2=(-7+-sqrt37)/2.#

# rArr t!=(-7+-sqrt37)/2, x'(t)!=0.#

#:., by (star), dy/dx={(1-3t)e^(-3t)}/{(1+7t+3t^2)e^t}, or, #

# dy/dx=(1-3t)/((1+7t+3t^2)e^(4t)), t!=(-7+-sqrt37)/2.#

Enjoy Maths.!