Radius

Posted By on March 28, 2014

Hello and welcome my dear reader, student or teacher. In this article we will discuss a brief introduction to the differential calculus. Coupled with the idea of integration, the derivation is an operation that focuses on taking parts of a whole. In the case that concerns us, the derivation, we want to take a tiny part of a very large whole. One of the most illustrative examples of bypass (and integration) are applied to the geometry derivatives. Imagine a sphere with RADIUS r, also suppose that it is a solid sphere.

Using the formula for the volume of the sphere we can know how many units occupy the space that encloses the sphere, but and if I just want to know how many units there are in the outer layer? Said in another way. Suppose we have a wooden ball and I want to know how much occupies only the varnish, a varnish as thin as we can imagine, indeed the finest varnish you can imagine. A problem of this nature is resolved in intuitive way by subtracting the volume of the original sphere a sphere a little (but (exactly only a little) smaller. Thus be empty and only remains a fine (the most thin layer possible) of elements that comprise it. Using bypass the problem can be resolved also, deriving the volume of a sphere can be to the surface.

Then, if I can do it by subtracting a slightly smaller sphere what is derivation useful in this case? Well, if it can be done by subtracting a small sphere it is because we know exactly its formula for calculating its volume, but if we don’t know it by its derivative we can know that value. Another way to view those derived using the same example of the areas is as a way to lower dimensions. We started again with a formula that speaks of one space in 3d, the formula for the volume of a sphere.

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