Abhisar
  • Abhisar
The relation between time t and distance x is \(\sf t = ax^² + bx\) where a and b are constants. The acceleration is
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Abhisar
  • Abhisar
I don't know how to do implicit differentiation.....
SolomonZelman
  • 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.
SolomonZelman
  • 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.

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SolomonZelman
  • SolomonZelman
Acceleration is a 2nd derivative (with respect to t)
Abhisar
  • Abhisar
Yep I know that but i was not able to differentiate it correctly, would you mind checking my work?
SolomonZelman
  • SolomonZelman
Yes, sure....
SolomonZelman
  • SolomonZelman
oh, lets hold a sec and re-write the function better....
Abhisar
  • Abhisar
Ok, So i was heading like this, differentiating the equation on both side by dt, \(\sf \frac{d}{dt}t= ax.\frac{dx}{dt}+b\frac{dx}{dt}\\ 1=axv + bv\)
SolomonZelman
  • SolomonZelman
and for even less confusion \(\large\color{black}{ \displaystyle x(t)=\frac{-b+\sqrt{b^2+4at}}{2a} }\) \(\large\color{black}{ \displaystyle x(t)=\frac{-b}{2a}+\frac{\sqrt{b^2+4at}}{2a} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x(t)=\frac{-b}{2a}+\frac{1}{2a}\cdot\sqrt{b^2+4at} }\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x'(t)=0+\frac{1}{2a}\cdot(b^2+4at)^{1/2~\color{red}{-1}} \color{red}{\times (1/2)}}\)
SolomonZelman
  • SolomonZelman
kind of like that, if it makes sense to you. (we can do a u-sub, if it is too messy for you)
SolomonZelman
  • SolomonZelman
oh, i frgot ththe chain
Abhisar
  • Abhisar
Yep I am getting it.
Abhisar
  • Abhisar
@SolomonZelman but can you check what i did and confirm if it is correct?
SolomonZelman
  • SolomonZelman
oh, sure.. let me look back..
Abhisar
  • Abhisar
Ok, So i was heading like this, differentiating the equation on both side by dt, \(\sf \frac{d}{dt}t= ax.\frac{dx}{dt}+b\frac{dx}{dt}\\ 1=axv + bv\)
SolomonZelman
  • SolomonZelman
yah that seems better than my way
SolomonZelman
  • SolomonZelman
((and your v is your velocity)) now, a second deriv. is going to be accelearation.
Abhisar
  • Abhisar
Yes, exactly but I am finding a different story in one of my book for this question. Lemme copy that for you.
Abhisar
  • Abhisar
Differentiate the equation on both sides wrt dt which will give \(\sf 1 = 2axv + bv\)
SolomonZelman
  • SolomonZelman
wrt?
SolomonZelman
  • SolomonZelman
with respect?
Abhisar
  • Abhisar
wrt = with respect to time
SolomonZelman
  • 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)
Abhisar
  • Abhisar
Ok but I did the same and got 1 = axv + bv
SolomonZelman
  • SolomonZelman
wait, you are saying that you got this for the 2nd deriv. or what ?
Abhisar
  • Abhisar
For first derivative I got 1 = axv + bv but the book says 1=2axv + bv. How?
SolomonZelman
  • SolomonZelman
oh, the power of 2 by x²
SolomonZelman
  • SolomonZelman
t=ax²+bx (d/dt) t= (d/dt) ax+ (d/dt) bx (dt/dt)= 2 (dx/dt) ax²⁻¹ + (dx/dt) b
SolomonZelman
  • 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
SolomonZelman
  • SolomonZelman
This is not just •C on every term in the equation. Playing with derivatives gets a little different.....
Abhisar
  • 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 ?
SolomonZelman
  • 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")
SolomonZelman
  • 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)
SolomonZelman
  • SolomonZelman
Want an implicit differention example, real quick?
Abhisar
  • Abhisar
Yes, I think that's my problem. I don't know how to such problems.
SolomonZelman
  • SolomonZelman
Lets do some implicit differentiation... it will be easy if you know basic properties of regular (not implicit differentiation).
Abhisar
  • Abhisar
Yes I know basics.
SolomonZelman
  • SolomonZelman
yeah, so i will do example or 2-3....
Abhisar
  • Abhisar
ok
SolomonZelman
  • 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'
SolomonZelman
  • SolomonZelman
sorry for long reply. (there is one step in implicit differentiation to solve for y' (just algebraic manipulations)
SolomonZelman
  • SolomonZelman
the important part is the chain rule of y' note. (take you time)
Abhisar
  • Abhisar
ok, so what is y'?
SolomonZelman
  • 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'
SolomonZelman
  • SolomonZelman
So basically y' is f'(x)
Abhisar
  • 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, \(\sf \frac{d}{dt}(t) = \frac{dx}{dt}.(2ax)+ b.\frac{dx}{dt}\)
Abhisar
  • Abhisar
\(\sf \Rightarrow 1 = 2axV + bV)\)
SolomonZelman
  • SolomonZelman
yes
Abhisar
  • Abhisar
Now, 1/v = 2ax + b Differentiating it one more time \(\sf -v^{-2}.a = 2a.V + 0 \\ \Rightarrow a = -2av^3\)
SolomonZelman
  • SolomonZelman
why do you have .a on the left side? and with respect to what are you differentiating?
Abhisar
  • Abhisar
With time
Abhisar
  • Abhisar
dv/dt = a (acceleration)
Abhisar
  • Abhisar
The constant a on the right hand side is different..
Abhisar
  • Abhisar
Hehe, I think that's correct.
SolomonZelman
  • 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, \(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle A=\frac{2a}{\sqrt{(b^2+4at)^3}}} }\)
SolomonZelman
  • 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)
SolomonZelman
  • SolomonZelman
A is the acceleration
Abhisar
  • Abhisar
Ummm.. I have four options 2av^2 2av^3 2abv^3 2b^2v^3
Abhisar
  • Abhisar
Which means i need to find the value in terms of v (velocity)
SolomonZelman
  • SolomonZelman
yeah...
SolomonZelman
  • 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
SolomonZelman
  • SolomonZelman
but I got negative same thing...
Abhisar
  • Abhisar
Yes that's correct. In option it's positive because they are asking Retarding force.
Abhisar
  • Abhisar
Thanks for bearing with me c:
SolomonZelman
  • 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!
Abhisar
  • Abhisar
Lel, thank you for time and help, you were life saver today ;)
SolomonZelman
  • SolomonZelman
Okay... good luck

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