The relation between time t and distance x is \(\sf t = ax^² + bx\) where a and b are constants. The acceleration is

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

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