Substitution easy question

Hello, could anyone help me a bit? I don't know how I could determine that g' measures how much a small interval shrinks or stretches from that graph. And the step in derivation marked with red arrow.

Note by Nezajem123 Nezajem123
1 year ago

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Comments

Hi Nezajem123 Nezajem123, I believe you're referring to this problem.

the reasoning goes like this:

The (infinitesimal) change in xx (Δx\Delta x) is the difference between the images of the corresponding endpoints of the interval along the uu direction under gg. In short, Δx=g(u1)g(u0)\Delta x = g(u_1)-g(u_0) with Δu=u1u0\Delta u = u_1-u_0.

We want to get a relationship between Δx\Delta x and Δu\Delta u from this expression. We start by writing u1u_1 as u0+Δuu_0+ \Delta u, which implies Δx=g(u0+Δu)g(u0)\Delta x = g(u_0+\Delta u) - g(u_0). Because Δu\Delta u is so very small, we can use a linear approximation to replace g(u0+Δu)g(u_0 +\Delta u ) by g(u0)+g(u0)Δug(u_0) + g’(u_0) \Delta u. This implies that Δxg(u0)+g(u0)Δug(u0)\Delta x \approx g(u_0) + g’(u_0) \Delta u - g(u_0), and so Δxg(u0)Δu\Delta x \approx g’(u_0) \Delta u. Another way of expressing this is as Δx/Δu=g(u0)\Delta x/ \Delta u = g’(u_0), so g(u0)g’(u_0) measure how much gg stretches/shrinks small intervals along the uu-direction into small intervals in the xx-direction.

Hope that helps!

Brilliant Mathematics Staff - 1 year ago

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Thank you very much. The step I didn't catch is linear approximation, another quiz solved it for me :) https://brilliant.org/practice/linear-approximation/?p=1

nezajem123 nezajem123 - 12 months ago
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