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Show that the curve with parametric equations $ x = t^2 $, $ y = 1 - 3t $, $ z = 1 + t^3 $ passes through the points $ (1, 4, 0) $ and $ (9, -8, 28) $ but not through the point $ (4, 7, -6) $.

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Calculus 3

Chapter 13

Vector Functions

Section 1

Vector Functions and Space Curves

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Lectures

03:04

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x. The input of a function is called the argument and the output is called the value. The set of all permitted inputs is called the domain of the function. Similarly, the set of all permissible outputs is called the codomain. The most common symbols used to represent functions in mathematics are f and g. The set of all possible values of a function is called the image of the function, while the set of all functions from a set "A" to a set "B" is called the set of "B"-valued functions or the function space "B"["A"].

08:32

In mathematics, vector calculus is an important part of differential geometry, together with differential topology and differential geometry. It is also a tool used in many parts of physics. It is a collection of techniques to describe and study the properties of vector fields. It is a broad and deep subject that involves many different mathematical techniques.

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Show that the curve with its parametric equations x, equal to t square y, is equal to 1. Minus 3 to 3 is equal to 1 plus. This cube passes through the points 140 and 9 negative 828, but not through the .4000 negative 6 point. So first we like to y is equal to 4, so we have 1. Minus 3 t is equal to 4, so we have t, is equal to negative 1 and when t is equal to 1 negative 1 wet half x is equal to 1. Z is equal to 0, so the .140 plus on the curve, and then we lit when y is equal to negative 8. So behalf 1, minus 3 t is equal to negative 8 point. So 3 t is equal to 9. T is equal to 3 and when t is equal to 3 x is equal to 9 and z is equal to 1 plus 3 q. So this is equal to 28 point for the .9 negative 828 lies on the curve, and then we light y is equal to 7 point. So we have 1. Minus 3 t is equal to 7. T is equal to negative 2 and 1 t is equal to negative 2 by half x is equal to t square, so this is equal to 4 and z is equal to 1 minus 8. So this is equal to negative 7 to the .747 negative 7 lies under curve. Here. Distance is negative 6, so the .47 negative 6 doesn't lie on the curve. Since, when y equal to 7, we half z is equal to negative 7.

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