## anonymous one year ago Given the functions x[n] = 1.3n + Sin[n] y[n] = 2(1 + Cos[n/2]) Assuming that these two functions are used to create a parametric plot where x = x[n] and y=y[n] Ah.. but lets say that x[n] is.. x[Pi/2] = (1.3(Pi))/2 + 1 = 3.0420352248333655​ How would you combine these two functions to find the derivative of y= f[x]? so you could work out the slope at that particular point?

1. anonymous

Why I am seeing this post as yellow colored? Or anybody else seeing the same?

2. anonymous

changed it a bit..had it wrong at first. I dont think finding an f[x] is very doable

3. anonymous

its a paid question to the qualified helpers.. they go yellow

4. ganeshie8

You may work the derivative directly w/o eliminating the parameter : $\large \dfrac{dy}{dx} = \dfrac{~\dfrac{dy}{dn}~}{~\dfrac{dx}{dn}~}$ provided ofcourse $$\dfrac{dx}{dn}\ne 0$$

5. anonymous

thnx, I can probably work this out from here..

6. anonymous

still not intuitive to me yet though.. even though I've done a dozen problems like this already

7. ganeshie8

guess it takes some practice, derivation is pretty straightforward though : $y = f((x)$ plugin the parameterization and get $2(1+\cos(n/2)) = f\left(x\right)$ differentiating with respect to $$n$$ both sides gives $-\sin(n/2) = f'\left(x\right)*x' = f'(x)*(1.3+\cos(n))$ $\implies f'\left(x\right) = \dfrac{-\sin(n/2)}{1.3+\cos(n)}$ which is same as what you would get using the earlier formula

8. anonymous

ahh, I had got this far.. had it upside down dx/dy = 1.3 + Cos[n] / -Sin[n/2]

9. ganeshie8

Looks good! simply flip both sides to get dy/dx you can play with derivatives just as fractions in single variable calculus

10. anonymous

ah gotcha, thank you lots, I couldn't see how to approach this problem at all

11. anonymous

Ive worked out lots of single function derivatives, but this is my first 2 function derivative

12. ganeshie8

You will see more of these in polar coordinates... If you take calcIII, you get lot of practice while parameterizing for line integrals and surfaces etc. Just want you know that parametric equations are very common and important!

13. anonymous

I am not a qualified helper..

14. anonymous

Have you tagged me somewhere writing this question?

15. anonymous

thank you, ganshie, I'll get this down..found a chapter on predator prey models, looks like it will clear it up for me, well and good..

16. anonymous

waterineyes.. not tagged mate.. it shows yellow / orange color to all members..

17. anonymous

The question in which you need a Qualified Helper, that question goes Yellow for all the users?

18. UsukiDoll

@waterineyes yes. OpenStudy has just added Qualified Helpers and OwlBucks which is their currency used to pay for immediate assistance. You can buy OwlBucks with PayPal or any major credit cards starting at \$1.99 for 10 OwlBucks

19. UsukiDoll