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January 31st, 2016, 06:38 PM   #1
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Chain rule to find dy/dt

Hello again,

I am stuck at the question below and can't make out what exactly I need to do.

I have been asked to use the Leibnitz notation of the chain rule to find dy/dt at t = −1 for
y = x³ − 5x² and x =√ (t² + 3)

Am I to somehow combine y and x and then substitute t? Would anyone be kind enough to clarify this for me?

Last edited by skipjack; February 2nd, 2016 at 04:02 AM. Reason: to insert missing parentheses
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January 31st, 2016, 07:25 PM   #2
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dy/dt = (dy/dx)(dx/dt)
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January 31st, 2016, 08:09 PM   #3
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So here is how far I got:

dy/dx = 3x²-10x
dx/dt = (t²+3)¹⁄²
= t(t2+3)⁻¹⁄²
at t=-1 => -1/2

What do I do next, to find dy/dt, am I to multiply (3x²-10x)( -1/2)?Am I on the right track?
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January 31st, 2016, 09:02 PM   #4
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You are on the right track. Your next step should be determine the value of $x$ when $t=-1$ and put that into your expression.
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January 31st, 2016, 10:00 PM   #5
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Quote:
Originally Posted by v8archie View Post
You are on the right track. Your next step should be determine the value of $x$ when $t=-1$ and put that into your expression.
Determine the value of x how?
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January 31st, 2016, 10:10 PM   #6
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You gave an expression for $x$ in terms of $t$ in your first post.
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January 31st, 2016, 10:15 PM   #7
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Quote:
Originally Posted by v8archie View Post
You gave an expression for $x$ in terms of $t$ in your first post.
If I substitute t=-1 for x I get ( -1/2)
Then to find dy/dt do I simply miltiply (3x²-10x)( -1/2)?
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February 1st, 2016, 09:38 AM   #8
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Quote:
Originally Posted by atari View Post
So here is how far I got:

dy/dx = 3x²-10x
dx/dt = (t²+3)¹⁄²
You mean x= (t²+3)¹⁄², not dx/dt.
Quote:
= t(t2+3)⁻¹⁄²
This is dx/dt

Quote:
at t=-1 => -1/2

What do I do next, to find dy/dt, am I to multiply (3x²-10x)( -1/2)?Am I on the right track?
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February 1st, 2016, 01:35 PM   #9
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$\displaystyle x=\sqrt{t^2+3}$
At t=-1, x=2.
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February 1st, 2016, 03:47 PM   #10
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I guess got this wrong then. This was a hard concept to grasp.

The answer I submitted was this:

dy/dx = 3x^2-10x
dx/dt = t(t^2+3)^-1/2

dy/dt = (3x^2-10x)(t/sqrtt^2+3)
substituting t=-1
(3x^2-10x)(-1/2)
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