Chapter 13: Problem 35
Show that the curve with parametric equations \(x=t^{2}\) , \(y=1-3 t, z=1+t^{3}\) passes through the points \((1,4,0)\) and \((9,-8,28)\) but not through the point \((4,7,-6)\)
Short Answer
Expert verified
The curve passes through (1,4,0) and (9,-8,28), but not through (4,7,-6).
Step by step solution
01
Determine t for Point (1,4,0)
To check if the curve passes through the point \((1,4,0)\), equate the parametric equations to the point. Start with \(x = t^2 = 1\). Solving for \(t\) gives us \(t = \pm 1\). Next, we check these values in the other two equations: \(y = 1 - 3t\) and \(z = 1 + t^3\).
02
Verify Point (1,4,0) for t=1
Substitute \(t = 1\) into the equations. \(y = 1 - 3(1) = -2\) and \(z = 1 + (1)^3 = 2\). Since \( (1, -2, 2)\) does not match \((1, 4, 0)\), we discard \(t = 1\).
03
Verify Point (1,4,0) for t=-1
Substitute \(t = -1\) into the equations. \(y = 1 - 3(-1) = 4\) and \(z = 1 + (-1)^3 = 0\). Thus, \((1, 4, 0)\) matches the expected coordinates for \(t = -1\). This point is on the curve.
04
Determine t for Point (9,-8,28)
For the point \((9, -8, 28)\), set \(x = t^2 = 9\). Solving gives \(t = \pm 3\). Check these values in the equations for \(y\) and \(z\).
05
Verify Point (9,-8,28) for t=3
For \(t = 3\), substitute into the equations: \(y = 1 - 3(3) = -8\) and \(z = 1 + (3)^3 = 28\). The coordinates match \((9, -8, 28)\), so the point is on the curve.
06
Verify Point (9,-8,28) for t=-3
For \(t = -3\), substitute into the equations: \(y = 1 - 3(-3) = 10\) and \(z = 1 + (-3)^3 = -26\). Since \((9, 10, -26)\) does not match the desired point, discard \(t = -3\).
07
Determine t for Point (4,7,-6)
For the point \((4, 7, -6)\), start with \(x = t^2 = 4\), giving \(t = \pm 2\). Check these in the equations for \(y\) and \(z\).
08
Verify Point (4,7,-6) for t=2
For \(t = 2\), substitute into the equations: \(y = 1 - 3(2) = -5\) and \(z = 1 + (2)^3 = 9\). The coordinates \((4, -5, 9)\) do not match \((4, 7, -6)\).
09
Verify Point (4,7,-6) for t=-2
For \(t = -2\), substitute into the equations: \(y = 1 - 3(-2) = 7\) and \(z = 1 + (-2)^3 = -7\). While the value for \(y\) matches, \(z = -7 eq -6\), so the point is not on the curve.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Calculus
Vector calculus is a mathematical tool used to analyze and describe vectors and scalar fields. In this exercise, we deal with a curve in three-dimensional space defined by parametric equations. These equations use a parameter, usually denoted as \(t\), to express the \(x\), \(y\), and \(z\) coordinates of points on a curve. This method is powerful because:
- It unambiguously defines each point along the curve.
- Allows easy calculations of derivatives and integrals, important for understanding the behavior of the curve.
Curve
A curve, in mathematical terms, is a continuous mapping from a parameter space (usually a time-like interval) to geometric space. Here, the curve is described by the parametric equations:
- \(x(t) = t^2\)
- \(y(t) = 1 - 3t\)
- \(z(t) = 1 + t^3\)
- Describing complex shapes that cannot be expressed as the graph of a function of \(x\) or \(y\) alone.
- Studying the orientation and direction of the curve as \(t\) varies.
Coordinate Geometry
Coordinate geometry, or analytic geometry, involves the study of geometry using a coordinate system. In the current exercise, coordinate geometry integrates with vector calculus to determine if given points lie on a specified curve. Using coordinates allows us to:
- Plot points in a space with respect to \(x, y, z\) axes.
- Measure distances and angles between points conveniently.
- Apply algebraic methods to solve geometric problems.
3D Space
Three-dimensional space is a geometric setting in which three values (or coordinates) are required to determine the position of an element (i.e., a point). Our exercise considers points defined in 3D with coordinates \((x, y, z)\). Key points about 3D space and parametric equations include:
- 3D space models real-world phenomena accurately since it accommodates width, height, and depth.
- Lines or curves can indicate motion and trajectory, essential for fields like physics and engineering.
- The use of parametric equations in 3D allows seamless calculation of intersections or points of contact like those in this exercise.