Problem 17
Question
Find an equation in \(x\) and \(y\) whose graph contains the points on the curve \(C\). Sketch the graph of \(C\), and indicate the orientation. $$\boldsymbol{x}=t^{2}, \quad \quad y=2 \ln t, \quad t>0$$
Step-by-Step Solution
Verified Answer
The equation is \(y = \ln x\). The graph is a rightward-increasing curve.
1Step 1: Express t in terms of x
The equations of the curve are given as \( x = t^2 \) and \( y = 2 \ln t \). From the first equation, express \( t \) in terms of \( x \): \( t = \sqrt{x} \). Since \( t > 0 \), we have \( t = \sqrt{x} \).
2Step 2: Substitute into y equation
Substitute \( t = \sqrt{x} \) into the equation for \( y \). We get \( y = 2 \ln(\sqrt{x}) \).
3Step 3: Simplify the equation
Using the logarithmic identity \( \ln(a^b) = b\ln(a) \), simplify the expression: \( \ln(\sqrt{x}) = \ln(x^{1/2}) = \frac{1}{2}\ln(x) \). Thus, \( y = 2 \times \frac{1}{2}\ln(x) = \ln(x) \).
4Step 4: Write the equation of the curve
The simplified equation representing the curve \( C \) is \( y = \ln x \).
5Step 5: Sketch the graph and indicate orientation
The graph of \( y = \ln x \) is a curve starting near the y-axis, rising to the right and increasing slowly. The orientation of the curve, as \( t \) increases from just above 0, is from left to right.
Key Concepts
Logarithmic FunctionsCurve SketchingGraph Orientation
Logarithmic Functions
Logarithmic functions are a crucial part of mathematics, providing a way to solve equations involving exponential growth or decay. In this exercise, we encounter the logarithm function in the form of the natural logarithm, denoted as \( \ln \). The natural logarithm has the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718.
Logarithmic functions have particular properties that make them useful in simplifying complex expressions. One such property is \( \ln(a^b) = b\ln(a) \), which allows us to break down the logarithm of an exponent into a simpler multiplied form.
This concept is used in the problem solution. When substituting \( t = \sqrt{x} \) into the y-equation, the expression \( \ln(\sqrt{x}) \) is simplified using the logarithmic property to become \( \frac{1}{2}\ln(x) \). This step is key in reaching the neat equation \( y = \ln x \) for the curve. By understanding logarithmic rules like this, students can handle more intricate equations and transformations.
Logarithmic functions have particular properties that make them useful in simplifying complex expressions. One such property is \( \ln(a^b) = b\ln(a) \), which allows us to break down the logarithm of an exponent into a simpler multiplied form.
This concept is used in the problem solution. When substituting \( t = \sqrt{x} \) into the y-equation, the expression \( \ln(\sqrt{x}) \) is simplified using the logarithmic property to become \( \frac{1}{2}\ln(x) \). This step is key in reaching the neat equation \( y = \ln x \) for the curve. By understanding logarithmic rules like this, students can handle more intricate equations and transformations.
Curve Sketching
Curve sketching involves plotting a graph of a function by identifying its important characteristics, such as intercepts, slope, and curvature.
With the curve \( y = \ln x \), students should note that:
Clarity in drawing these characteristics will aid in understanding how the function behaves. Visualizing this gradual increase and asymptotic nature will enhance skills in representing complex mathematical relationships graphically.
With the curve \( y = \ln x \), students should note that:
- There is no y-intercept, as the curve approaches the y-axis but does not touch or include \( x = 0 \).
- The function is defined only for \( x > 0 \).
- The curve increases slowly as \( x \) becomes larger, giving it a gentle upward slope.
- It never touches the x-axis but gets closer as \( x \) becomes infinitely large.
Clarity in drawing these characteristics will aid in understanding how the function behaves. Visualizing this gradual increase and asymptotic nature will enhance skills in representing complex mathematical relationships graphically.
Graph Orientation
Graph orientation indicates the direction in which a parametric curve is traversed, often dictated by an underlying parameter like \( t \) in our case.
For the parametric equations \( x = t^2 \) and \( y = 2 \ln t \), both \( x \) and \( y \) are expressed in terms of \( t \), where \( t > 0 \). As \( t \) increases from just above 0, \( x = t^2 \) grows larger, moving us from left to right on the graph. This manifests the orientation of the curve, as seen while sketching \( y = \ln x \).
Understanding orientation helps in correctly plotting parametric curves, as the parameter's progression defines the path along the graph. Students should pay attention to initial conditions, like \( t \) being positive, which determine where the curve begins and how it proceeds. Mastering these concepts will lay a solid foundation for exploring more sophisticated parametric and vector functions.
For the parametric equations \( x = t^2 \) and \( y = 2 \ln t \), both \( x \) and \( y \) are expressed in terms of \( t \), where \( t > 0 \). As \( t \) increases from just above 0, \( x = t^2 \) grows larger, moving us from left to right on the graph. This manifests the orientation of the curve, as seen while sketching \( y = \ln x \).
Understanding orientation helps in correctly plotting parametric curves, as the parameter's progression defines the path along the graph. Students should pay attention to initial conditions, like \( t \) being positive, which determine where the curve begins and how it proceeds. Mastering these concepts will lay a solid foundation for exploring more sophisticated parametric and vector functions.
Other exercises in this chapter
Problem 16
Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$25 x^{2}-9 y^{2}+100
View solution Problem 16
Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$x=5$$
View solution Problem 17
Find a polar equation that has the same graph as the equation in \(x\) and \(y\). $$x^{2}+y^{2}=16$$
View solution Problem 18
Find an equation in \(x\) and \(y\) whose graph contains the points on the curve \(C\). Sketch the graph of \(C\), and indicate the orientation. $$x=\cos ^{3} t
View solution