- #1

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- Thread starter MRAI
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- #1

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- #2

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Then set t=1 in the derivative.

- #3

- 127

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I'll help you with the first part of the derivative. Remember that you can take the derivative of a sum piece by piece.

So the first part is [tex]4x^2= f(x(t) )[/tex] where f in this case is [tex]f(x) = 4x^2[/tex] and [tex]f'(x)=8x[/tex]. So from the chain rule [tex]\frac{d}{dt} f(x(t))= f'(x(t))\cdot x'(t) = 8x(t) * x'(t)[/tex]

Then once you have everything you just need to figure out what x(1), x'(1), y(1), y'(1) are and plug in. Easy

This way really does lead to a lot less work than plugging in, expanding, then differentiating.

- #4

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If yes, then how u find dz/dt in terms of t?

- #5

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i still VERY confused, about the chain rule! :(

- #6

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All the chain rule says is if a function can just be written as a function of another function then there is a short cut. Do you see that if f is defined as [tex]f(x) = 4x^2[/tex] then [tex]f(t+3) = 4(t+3)^2[/tex]? Don't be embarresed if that isn't clear... I've seen many calc students not understand how basic functions work but you will need to figure it out to get through the class.

Here's a bit more help define f as [tex]f(x) = 4x^2 +6x -2[/tex] and g as [tex]g(x) = -x^2[/tex]. Now write out f( x(t) ) + g( y(t) ). What is that (hint look at the original problem)? Given that can you solve [tex]\frac{d}{dt} f( x(t) ) [/tex] from what I've told you before? What about [tex]\frac{d}{dt} g(y(t)) [/tex]. Then what about [tex]\frac{d}{dt}(f(x(t)) + g(y(t)) )[/tex]

As I say if this stuff is too confusing don't use it. There is always expanding the polynomial and taking the derivative that way.

Good luck

Steven

- #7

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[tex] \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt} [/tex]

It's simple.Compute the 4 differentiations and express everything in terms of 't'.

Daniel.

- #8

- 70

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dz/dt=8x(dx/dt)-6y(dy/dt)+6(dx/dt),where dx/dt=1,and dy/dt=4t^3-2t^2.now substitute these values and then put t as i,u get dz/dt to be 38.see ya

- #9

uart

Science Advisor

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dextercioby said:

[tex] \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt} [/tex]

It's simple.Compute the 4 differentiations and express everything in terms of 't'.

Daniel.

Yes indeed, it would appear that this is an exercize aimed squarely at using the chain rule as posted above.

MRAI, it you're just learning about this stuff I recommend that you do it both ways as an exercize. Firstly do it using the chain rule as posted above by Daniel and then repeat using simple substitution. If you make no errors then you will certainly get the same result either way.

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