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

- Abhisar

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- Abhisar

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

- 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

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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## More answers

- SolomonZelman

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

- Abhisar

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

- SolomonZelman

Yes, sure....

- SolomonZelman

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

- 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

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

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

- 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

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

- SolomonZelman

oh, i frgot ththe chain

- Abhisar

Yep I am getting it.

- Abhisar

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

- SolomonZelman

oh, sure.. let me look back..

- Abhisar

- SolomonZelman

yah that seems better than my way

- SolomonZelman

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

- Abhisar

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

- Abhisar

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

- SolomonZelman

wrt?

- SolomonZelman

with respect?

- Abhisar

wrt = with respect to time

- 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

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

- SolomonZelman

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

- Abhisar

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

- SolomonZelman

oh, the power of 2 by x²

- SolomonZelman

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

- 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

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

- 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

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

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

Want an implicit differention example, real quick?

- Abhisar

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

- SolomonZelman

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

- Abhisar

Yes I know basics.

- SolomonZelman

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

- Abhisar

ok

- 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

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

- SolomonZelman

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

- Abhisar

ok, so what is y'?

- 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

So basically y' is f'(x)

- 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

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

- SolomonZelman

yes

- Abhisar

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

- SolomonZelman

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

- Abhisar

With time

- Abhisar

dv/dt = a (acceleration)

- Abhisar

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

- Abhisar

Hehe, I think that's correct.

- 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

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

A is the acceleration

- Abhisar

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

- Abhisar

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

- SolomonZelman

yeah...

- 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

but I got negative same thing...

- Abhisar

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

- Abhisar

Thanks for bearing with me c:

- 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

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

- SolomonZelman

Okay... good luck

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