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#1 2006-04-14 04:18:36

dadon
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Rate of change

How do you do the following rate of change questions

when t = 1.2 for all the following questions:

1) y = 4/t + 4lnt

2) x = sin(t^2 +1)  ,  y = cos(2t -3)

3) x = 1 / 1+2t  ,  y = t / 1+t

4) q = 2e^-t/2.cos2t       (e is exponential for this question)

5) x = e^2t.t^3(2-t)^4   (e is exponential for this question)


If someone could help I would be very greatful!

Thanks

#2 2006-04-15 09:40:00

yonski
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Re: Rate of change

You need to calculate the derivative of each function...

1) d/dt(4/t +4lnt) = -4/t^2 + 4/t

2) d/dt(sin(t^2 + 1)) = 2cos(t^2 + 1)t

    d/dt(cos(2t - 3)) = -2sin(2t - 3)            (both using the composite rule)

3) d/dt(2(e^(-t/2))cos(2t)) = (-e^(-t/2))cos(2t) - 4(e^(-t/2))sin(2t)       (using the product rule)

4) I'm not sure about this one, it's a little bit unclear what you mean, but you get the general idea so far. Hopefully the rest are okay, but it's late so i'd check em! smile

Oh math, i've just realised you've specified t=1.2, so you'd have to use that value with the derived functions to get your answers.

Last edited by yonski (2006-04-15 09:48:41)


Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

#3 2006-04-15 21:31:32

dadon
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Re: Rate of change

Heys thanks I get what you have done!

Could you explain the composite rule as I havn't come across that.

Thanks agian!smile

#4 2006-04-15 23:10:33

yonski
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Re: Rate of change

Composite Rule (sometimes called the Chain Rule)

If k is a function with rule of the form k(x) = g(f(x)), where f and g are smooth functions, then k is smooth and

      k'(x) = g'(f(x)).f'(x)

(or using Leibniz form:  If y = g(u), where u = f(x), then dy/dx = (dy/du)(du/dx) ) .


So, for instance, if you have k(x) = ln(x^2 + 1), then y = g(u), where g(u) = lnu and u = x^2 + 1. This gives k'(x) = (1/(x^2 + 1)).2x = 2x/(x^2 + 1) .

Hope that helps.


Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

#5 2006-04-15 23:37:22

dadon
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Re: Rate of change

Thanks its just the chain rule! cool

At least I don't have to learn anything new!

Regards,

dadon

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