Answers>Maths>IB>Article

Given the parametric equations x = lnt+t and y = sint calculate d^2y/dx^2

First we can write d2y/dx2 as (d/dx)(dy/dx). Now we need to find dy/dx. This can be further written as (dy/dt)(dt/dx). These derivatives can be obtained from the given parametric equations:
 dx/dt = 1/t + 1, hence dt/dx = 1/(1/t+1) = t/(1+t)
dy/dt = cost,
Therefore combining these to we obtain
dy/dx = tcost/(1+t).
Now goind back to what we wrote at the beginning - d2y/dx2 as (d/dx)(dy/dx) - we can write it as (dt/dx)(d/dx)(dy/dx) and this is equal to (dt/dx)(d/dt)[tcost/(t+1)].
First let's compute (d/dt)[tcost/(t+1)]:
Using the product and quotient rule (if we have to functions say f and g and their derivatives f' and g' product rule says that (fg)' = f'g + fg' and quotient rule says that (f/g) = (f'g-fg')/g2) we can write:
[tcost/(t+1)]' = (cost-tsint)(t+1)-tcost)/(t+1)2 = (cost-t2sint-tsint)/(t+1)2
Going bakc to the formula for the second derivative we need to multiply our result by dt/dx, therefore
  (cost-t2sint-tsint)/(t+1)2 (dt/dx) = (cost-t2sint-tsint)/(t+1)2 * [t/(1+t)] = t(cost-t2sint-tsint)/(t+1)3

Final answer: d2y/dx2 = t(cost-t2sint-tsint)/(t+1)3

AR
Answered by Agnieszka R. Maths tutor

3302 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

H(x)=(x^3)*(e^x) what is H'(x)


Differentiate, from first principles, y=x^2


integrate arcsin(x)


Find the coordinates that correspond to the maximum point of the following equation: y = −16x^2 + 160x - 256


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences