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Two particles travel along the space curves$ r_1 (t) = \langle t, t^2, t^3 \rangle $ $ r_2 (t) = \langle 1 + 2t, 1 + 6t, 1 + 14t \rangle $Do the particles collide? Do their paths intersect?
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does not collide, intersect at $(1,1,1)$ and $(2,4,8)$
Vector Functions and Space Curves
Johns Hopkins University
Oregon State University
Harvey Mudd College
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"].
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.
Two particles travel along…
If two objects travel thro…
The paths of two particles…
The problem is 2 particles. Travel along the space, curves were 1 t is t t square t, cube 2 t is equal to 1 plus 2 t 1 plus 61 plus 14 t do the particles plate to their pipes intersect. So first the particles collate means there exists some t 0 such that 10 is equal to 20 pont. So if they collide with half t 0 is equal to 1 plus 2 t 00 square is equal to 1 plus 6. T 0 is equal to 1 plus 140 point from the first equation. We have 0 is equal to negative 1 point. That 1 would plug in negative 1 to the second equation. We have 1 is equal to negative 5 point. This is a contradiction to the particles. Don'T collate with their positive intersect so that their positive intersect means there exists. T 1 and t 2 such that r 1 t 1 is equal to r 2 t 2. So behalf, if their positive half is intersect, we have t 1 square. T 1 is equal to 1 plus 2 times. T 2.1 square is equal to 1 plus 621 is equal to 1 plus 14. The first lot equation 1 and the equation 2 o 2 minus 1 is t 1. Squared minus t 1 is equal to 4 t 2 point and from equation 1 we have 42 is equal to 2 times. T 1 minus 1 point. So we solve these equations. We have t 1 is equal to 1 or 2 point when t 1 is equal to 1. T 2 is equal to 0 and when t 1 is equal to 2 t 2 is equal to half. Then we're plugging this 2 values to the equation. 3. We find in first case this is 1 equal to 1. This case 1 sequel 12, is equal to 0 point, so we have 1 equal to 1 and for this case we have 8 is equal to 8, so bossant true, then be half their passes intersect and they have 2 intersection points.
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