A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  2. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  5. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  6. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Yes, sure....

  7. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  8. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  11. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  12. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    oh, i frgot ththe chain

  14. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Yep I am getting it.

  15. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  16. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    oh, sure.. let me look back..

  17. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    yah that seems better than my way

  19. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  20. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  21. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  22. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    wrt?

  23. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    with respect?

  24. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    wrt = with respect to time

  25. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  27. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  28. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  29. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    oh, the power of 2 by x²

  30. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  31. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  33. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Want an implicit differention example, real quick?

  37. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  38. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  39. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Yes I know basics.

  40. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  41. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ok

  42. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  44. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  45. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    ok, so what is y'?

  46. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    So basically y' is f'(x)

  48. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  50. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    yes

  51. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  52. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

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

  53. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    With time

  54. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    dv/dt = a (acceleration)

  55. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  56. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Hehe, I think that's correct.

  57. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    A is the acceleration

  60. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  61. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  62. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    yeah...

  63. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    but I got negative same thing...

  65. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  66. Abhisar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Thanks for bearing with me c:

  67. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  69. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    Okay... good luck

  70. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.