## Abhisar one year ago The relation between time t and distance x is $$\sf t = ax^² + bx$$ where a and b are constants. The acceleration is

1. Abhisar

I don't know how to do implicit differentiation.....

2. SolomonZelman

This is kind of an implicit expression. $$\large\color{black}{ \displaystyle t=ax^2+bx }$$ I would first try to remodel it as a distance a function of time. $$\large\color{black}{ \displaystyle 0=ax^2+bx-t }$$ $$\large\color{black}{ \displaystyle x=\frac{-b\pm\sqrt{b^2+4at}}{2a} }$$ where x is distance and t is time.

3. SolomonZelman

So this is kind of a $$\large\color{black}{ \displaystyle x(t)=\frac{-b\pm\sqrt{b^2+4at}}{2a} }$$ where x(t) is a distance at a particular time t.

4. SolomonZelman

Acceleration is a 2nd derivative (with respect to t)

5. Abhisar

Yep I know that but i was not able to differentiate it correctly, would you mind checking my work?

6. SolomonZelman

Yes, sure....

7. SolomonZelman

oh, lets hold a sec and re-write the function better....

8. Abhisar

Ok, So i was heading like this, differentiating the equation on both side by dt, $$\sf \frac{d}{dt}t= ax.\frac{dx}{dt}+b\frac{dx}{dt}\\ 1=axv + bv$$

9. SolomonZelman

and for even less confusion $$\large\color{black}{ \displaystyle x(t)=\frac{-b+\sqrt{b^2+4at}}{2a} }$$ $$\large\color{black}{ \displaystyle x(t)=\frac{-b}{2a}+\frac{\sqrt{b^2+4at}}{2a} }$$

10. SolomonZelman

$$\large\color{black}{ \displaystyle x(t)=\frac{-b}{2a}+\frac{1}{2a}\cdot\sqrt{b^2+4at} }$$

11. SolomonZelman

$$\large\color{black}{ \displaystyle x'(t)=0+\frac{1}{2a}\cdot(b^2+4at)^{1/2~\color{red}{-1}} \color{red}{\times (1/2)}}$$

12. SolomonZelman

kind of like that, if it makes sense to you. (we can do a u-sub, if it is too messy for you)

13. SolomonZelman

oh, i frgot ththe chain

14. Abhisar

Yep I am getting it.

15. Abhisar

@SolomonZelman but can you check what i did and confirm if it is correct?

16. SolomonZelman

oh, sure.. let me look back..

17. Abhisar

Ok, So i was heading like this, differentiating the equation on both side by dt, $$\sf \frac{d}{dt}t= ax.\frac{dx}{dt}+b\frac{dx}{dt}\\ 1=axv + bv$$

18. SolomonZelman

yah that seems better than my way

19. SolomonZelman

((and your v is your velocity)) now, a second deriv. is going to be accelearation.

20. Abhisar

Yes, exactly but I am finding a different story in one of my book for this question. Lemme copy that for you.

21. Abhisar

Differentiate the equation on both sides wrt dt which will give $$\sf 1 = 2axv + bv$$

22. SolomonZelman

wrt?

23. SolomonZelman

with respect?

24. Abhisar

wrt = with respect to time

25. SolomonZelman

yeah that is what you should do, you differentiate with respect to time (or with respect to t, and your function is now velocity v, so you do dv/dt)

26. Abhisar

Ok but I did the same and got 1 = axv + bv

27. SolomonZelman

wait, you are saying that you got this for the 2nd deriv. or what ?

28. Abhisar

For first derivative I got 1 = axv + bv but the book says 1=2axv + bv. How?

29. SolomonZelman

oh, the power of 2 by x²

30. SolomonZelman

t=ax²+bx (d/dt) t= (d/dt) ax+ (d/dt) bx (dt/dt)= 2 (dx/dt) ax²⁻¹ + (dx/dt) b

31. SolomonZelman

you are differentiating after all when you mutliply times d/dt... so this is what you get, you would get a coefficient of 3 and a power of 2, if you have ax³, and you would get a coefficient of n, and power of n-1, if you had axⁿ

32. SolomonZelman

This is not just •C on every term in the equation. Playing with derivatives gets a little different.....

33. Abhisar

ok, so if i want to differentiate say $$\sf x^2$$ with respect to time I get d/dt $$\sf x^2$$ = 2x/dt = 2v ?

34. SolomonZelman

2x and times the derivative of x', if you are differentiating with respect to x. In this case this x' was the velocity (and namely, the dx/dt, or as you named it - which is typical - "v")

35. SolomonZelman

This is going to the implicit differentiation, that when you find dy/dx of a function in a form of f(x,y)=.... where the function is not defined explicity (has y's and x's mixed up - roughly speaking, (and that is why, as well) doesn't pass vertical line test)

36. SolomonZelman

Want an implicit differention example, real quick?

37. Abhisar

Yes, I think that's my problem. I don't know how to such problems.

38. SolomonZelman

Lets do some implicit differentiation... it will be easy if you know basic properties of regular (not implicit differentiation).

39. Abhisar

Yes I know basics.

40. SolomonZelman

yeah, so i will do example or 2-3....

41. Abhisar

ok

42. SolomonZelman

say i got: 4x+2y²x=4y (y is output, x is input) I differentiate all parts applying my rules, but... dy/dx (4x¹) = dy/dx (4x)=4x¹⁻¹ = 4x⁰ = 4 dy/dx (4y) = 4 • y' (same as by 4x, but we multiply times y', since y is a function of x.) Where does this come from? I will tell you. So say you wanted to take a derivative of a huge function f(x), you will have a chain rule - multiply times f'(x) (correct?) Here y' plays the same role. You are taking the derivative of this huge function f(x) (and y is here to denote a function f(x)), so just like by taking a derivative of a function f(x) you get a chain rule of f'(x), so is when you differentiate [a function] y, you get a chain rule of y'. dy/dx (2y²x) = x • dy/dx (2y²) + 2y² • dy/dx (x) (taking the product rule, and will apply the same chain rule principal by every derivative of y) it becomes: dy/dx (2y²x) = x • (2• 2y²⁻¹) • y' + 2y² • 1 see the y' here that i explained before? (and derivative of x, with respect to x, is just 1) simplifying: dy/dx (2y²x) = x(4y)y' + 2y² So overall, when we differentiate the: 4x+2y²x=4y (finding dy/dx) we are going to get: 4 + [ x(4y)y' + 2y² ] = 4 y'

43. SolomonZelman

sorry for long reply. (there is one step in implicit differentiation to solve for y' (just algebraic manipulations)

44. SolomonZelman

the important part is the chain rule of y' note. (take you time)

45. Abhisar

ok, so what is y'?

46. SolomonZelman

y' is the chain rule. y differens from x, in that y is [denoting] a function of x. here are some equivalent statements: dy/dx[ f(x) ] = f'(x) $$\Rightarrow$$ dy/dx[y]=y' dy/dx[ ( f(x) )³ ] = 3f(x)² • f'(x) $$\Rightarrow$$ dy/dx[y³]=3y²•y'

47. SolomonZelman

So basically y' is f'(x)

48. Abhisar

Ok, I think I got it. So Here I need to apply chain rule. For example if i want to differentiate the original equation, I will do it like this, $$\sf \frac{d}{dt}(t) = \frac{dx}{dt}.(2ax)+ b.\frac{dx}{dt}$$

49. Abhisar

$$\sf \Rightarrow 1 = 2axV + bV)$$

50. SolomonZelman

yes

51. Abhisar

Now, 1/v = 2ax + b Differentiating it one more time $$\sf -v^{-2}.a = 2a.V + 0 \\ \Rightarrow a = -2av^3$$

52. SolomonZelman

why do you have .a on the left side? and with respect to what are you differentiating?

53. Abhisar

With time

54. Abhisar

dv/dt = a (acceleration)

55. Abhisar

The constant a on the right hand side is different..

56. Abhisar

Hehe, I think that's correct.

57. SolomonZelman

lets see..... What i was gonna do from the beginning is: $$\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle x(t)=\frac{-b+\sqrt{b^2+4at}}{2a} } }$$ $$\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle x(t)=\frac{-b}{2a}+\frac{\sqrt{b^2+4at}}{2a} } }$$ $$\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle x'(t)=0+\frac{4a}{4a\sqrt{b^2+4at}} } }$$ $$\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle x'(t)=\frac{1}{\sqrt{b^2+4at}} } }$$ $$\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle x''(t)=\left(\sqrt{b^2+4at}\right)^{-1/2~~\color{red}{-1}} ~\cdot (1/2)4a} }$$ $$\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle x''(t)=2a\left(\sqrt{b^2+4at}\right)^{-3/2}} }$$ so, $$\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle A=\frac{2a}{\sqrt{(b^2+4at)^3}}} }$$

58. SolomonZelman

This is how I was going to do it, without any use of creativity or brain. Simply solve for x in terms of t, and find the derivative twice (both times doing dx/dt)

59. SolomonZelman

A is the acceleration

60. Abhisar

Ummm.. I have four options 2av^2 2av^3 2abv^3 2b^2v^3

61. Abhisar

Which means i need to find the value in terms of v (velocity)

62. SolomonZelman

yeah...

63. SolomonZelman

1=2axv+bv 1/v=2ax+b differentiating ... (dv/dt)1/v=(dx/dt)2ax+(d/dt)b -v$$^{-2}$$ • $$\color{red}{\rm A}$$ =2a•x' $$\color{red}{\rm A}$$ =-2v²a•x' this x' is v. $$\color{red}{\rm A}$$ =-2v²a•v $$\color{red}{\rm A}$$ =-2v³a

64. SolomonZelman

but I got negative same thing...

65. Abhisar

Yes that's correct. In option it's positive because they are asking Retarding force.

66. Abhisar

Thanks for bearing with me c:

67. SolomonZelman

Kind of an unusual for me. I would tend to find acceleration with respect to time (to know how particle accelerates at paricular time).... got it thorugh, tho'. next time won';t play basketball before going on here... hehe.. you are welcome, you were great!

68. Abhisar

Lel, thank you for time and help, you were life saver today ;)

69. SolomonZelman

Okay... good luck