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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
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

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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3I don't know how to do implicit differentiation.....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5This 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. \(\large\color{black}{ \displaystyle 0=ax^2+bxt }\) \(\large\color{black}{ \displaystyle x=\frac{b\pm\sqrt{b^2+4at}}{2a} }\) where x is distance and t is time.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5So 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.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5Acceleration is a 2nd derivative (with respect to t)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Yep I know that but i was not able to differentiate it correctly, would you mind checking my work?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5oh, lets hold a sec and rewrite the function better....

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.5and 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
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle x(t)=\frac{b}{2a}+\frac{1}{2a}\cdot\sqrt{b^2+4at} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle x'(t)=0+\frac{1}{2a}\cdot(b^2+4at)^{1/2~\color{red}{1}} \color{red}{\times (1/2)}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5kind of like that, if it makes sense to you. (we can do a usub, if it is too messy for you)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5oh, i frgot ththe chain

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3@SolomonZelman but can you check what i did and confirm if it is correct?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5oh, sure.. let me look back..

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.5yah that seems better than my way

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5((and your v is your velocity)) now, a second deriv. is going to be accelearation.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Yes, exactly but I am finding a different story in one of my book for this question. Lemme copy that for you.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Differentiate the equation on both sides wrt dt which will give \(\sf 1 = 2axv + bv\)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3wrt = with respect to time

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5yeah 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
 one year ago
Best ResponseYou've already chosen the best response.3Ok but I did the same and got 1 = axv + bv

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5wait, you are saying that you got this for the 2nd deriv. or what ?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3For first derivative I got 1 = axv + bv but the book says 1=2axv + bv. How?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5oh, the power of 2 by x²

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5t=ax²+bx (d/dt) t= (d/dt) ax+ (d/dt) bx (dt/dt)= 2 (dx/dt) ax²⁻¹ + (dx/dt) b

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5you 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 n1, if you had axⁿ

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5This is not just •C on every term in the equation. Playing with derivatives gets a little different.....

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3ok, 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
 one year ago
Best ResponseYou've already chosen the best response.52x 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
 one year ago
Best ResponseYou've already chosen the best response.5This 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
 one year ago
Best ResponseYou've already chosen the best response.5Want an implicit differention example, real quick?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Yes, I think that's my problem. I don't know how to such problems.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5Lets do some implicit differentiation... it will be easy if you know basic properties of regular (not implicit differentiation).

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5yeah, so i will do example or 23....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5say 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
 one year ago
Best ResponseYou've already chosen the best response.5sorry for long reply. (there is one step in implicit differentiation to solve for y' (just algebraic manipulations)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5the important part is the chain rule of y' note. (take you time)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5y' 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
 one year ago
Best ResponseYou've already chosen the best response.5So basically y' is f'(x)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.3\(\sf \Rightarrow 1 = 2axV + bV)\)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Now, 1/v = 2ax + b Differentiating it one more time \(\sf v^{2}.a = 2a.V + 0 \\ \Rightarrow a = 2av^3\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5why do you have .a on the left side? and with respect to what are you differentiating?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3dv/dt = a (acceleration)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3The constant a on the right hand side is different..

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Hehe, I think that's correct.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5lets 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
 one year ago
Best ResponseYou've already chosen the best response.5This 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
 one year ago
Best ResponseYou've already chosen the best response.5A is the acceleration

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Ummm.. I have four options 2av^2 2av^3 2abv^3 2b^2v^3

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Which means i need to find the value in terms of v (velocity)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.51=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
 one year ago
Best ResponseYou've already chosen the best response.5but I got negative same thing...

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Yes that's correct. In option it's positive because they are asking Retarding force.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.3Thanks for bearing with me c:

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5Kind 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
 one year ago
Best ResponseYou've already chosen the best response.3Lel, thank you for time and help, you were life saver today ;)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5Okay... good luck
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