Question

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

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

Yes, sure....

Answer 7

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

Answer 8

Ok,
So i was heading like this, differentiating the equation on both side by dt,

\(\sf \frac{d}{dt}t= ax.\frac{dx}{dt}+b\frac{dx}{dt}\\ 1=axv + bv\)

Answer 9

and for even less confusion

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

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

oh, i frgot ththe chain

Answer 14

Yep I am getting it.

Answer 15

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

Answer 16

oh, sure.. let me look back..

Answer 17

Ok,
So i was heading like this, differentiating the equation on both side by dt,

\(\sf \frac{d}{dt}t= ax.\frac{dx}{dt}+b\frac{dx}{dt}\\ 1=axv + bv\)

Answer 18

yah that seems better than my way

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

wrt?

Answer 23

with respect?

Answer 24

wrt = with respect to time

Answer 25

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)

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

oh, the power of 2 by x²

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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 ?

Answer 34

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")

Answer 35

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)

Answer 36

Want an implicit differention example, real quick?

Answer 37

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

Answer 38

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

Answer 39

Yes I know basics.

Answer 40

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

Answer 41

ok

Answer 42

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'

Answer 43

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

Answer 44

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

Answer 45

ok, so what is y'?

Answer 46

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'

Answer 47

So basically y' is f'(x)

Answer 48

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,

\(\sf \frac{d}{dt}(t) = \frac{dx}{dt}.(2ax)+ b.\frac{dx}{dt}\)

Answer 49

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

Answer 50

yes

Answer 51

Now, 1/v = 2ax + b

Differentiating it one more time

\(\sf -v^{-2}.a = 2a.V + 0 \\ \Rightarrow a = -2av^3\)

Answer 52

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

Answer 53

With time

Answer 54

dv/dt = a (acceleration)

Answer 55

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

Answer 56

Hehe, I think that's correct.

Answer 57

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,

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle A=\frac{2a}{\sqrt{(b^2+4at)^3}}} }\)

Answer 58

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)

Answer 59

A is the acceleration

Answer 60

Ummm.. I have four options

2av^2
2av^3
2abv^3
2b^2v^3

Answer 61

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

Answer 62

yeah...

Answer 63

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

Answer 64

but I got negative same thing...

Answer 65

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

Answer 66

Thanks for bearing with me c:

Answer 67

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!

Answer 68

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

Answer 69

Okay... good luck