Problem 39
Question
On the curve \(x^{3}=12 y\), the abscissa changes at a faster rate than the ordinate. Then, \(x\) belongs to the interval (A) \((-4,4)\) (B) \((-3,3)\) (C) \((-2,2)\) (D) None of these
Step-by-Step Solution
Verified Answer
The interval for \(x\) is \((-2, 2)\) (Option C).
1Step 1: Equation of the curve
The equation of the curve given is \(x^3 = 12y\). Here, \(x\) is the abscissa (horizontal coordinate), and \(y\) is the ordinate (vertical coordinate).
2Step 2: Differentiate to find rates of change
Differentiate the equation \(x^3 = 12y\) with respect to \(t\) (time) to find the relationship between the rates of change of \(x\) and \(y\). This gives us: \(3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}\).
3Step 3: Express the rates of change ratio
Rearrange the differentiated equation to express \(\frac{dx}{dt}\) in terms of \(\frac{dy}{dt}\): \(\frac{dx}{dt} = \frac{4 \frac{dy}{dt}}{x^2}\).
4Step 4: Establish inequality for faster abscissa change
Since the abscissa changes at a faster rate than the ordinate, we need \(\left| \frac{dx}{dt} \right| > \left| \frac{dy}{dt} \right|\). Substitute the expression and simplify: \(\left| \frac{4 \frac{dy}{dt}}{x^2} \right| > \left| \frac{dy}{dt} \right|\).
5Step 5: Solve the inequality
Cancel \(\left| \frac{dy}{dt} \right|\) from both sides (assuming it is not zero and has the same sign), thus \(\left| \frac{4}{x^2} \right| > 1\), leading to \(\frac{4}{x^2} > 1\).
6Step 6: Simplify and find interval for \(x\)
Solving the inequality \(\frac{4}{x^2} > 1\) gives \(x^2 < 4\). Taking square roots: \(-2 < x < 2\). So \(x\) belongs to the interval \((-2, 2)\).
Key Concepts
Curve AnalysisRate of ChangeAbscissa and Ordinate
Curve Analysis
Understanding curve analysis involves studying the shape, behavior, and properties of curves represented by algebraic equations. In our example, the curve is represented by the equation \(x^3 = 12y\). Here, the shape of the curve is determined by the relationship between the variables \(x\) and \(y\).
This curve is a cubic equation in \(x\), which means it can have various interesting features like points of inflection or can represent loops and bends. To analyze this curve, we often differentiate the equation to find slopes and study how rapidly different parameters change. Differentiating helps in understanding the dynamics and nature of how each variable changes relative to others, especially over time or space.
This curve is a cubic equation in \(x\), which means it can have various interesting features like points of inflection or can represent loops and bends. To analyze this curve, we often differentiate the equation to find slopes and study how rapidly different parameters change. Differentiating helps in understanding the dynamics and nature of how each variable changes relative to others, especially over time or space.
- Equation: The given equation \(x^3 = 12y\) directly links the horizontal position (or abscissa) \(x\) to the vertical position (or ordinate) \(y\).
- Applications: Curve analysis is crucial in many fields like physics, economics, and engineering, where we need to predict behavior based on current patterns.
Rate of Change
The concept of rate of change is crucial in differential calculus as it measures how a quantity changes over time or another variable. In this context, we consider how quickly the abscissa \(x\) and the ordinate \(y\) change with respect to time. Using calculus, we differentiate the equation \(x^3 = 12y\) with respect to time \(t\).
By differentiating, we find the rates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), representing the speeds of change of \(x\) and \(y\) respectively. The relationship derived from this differentiation is given by \(3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}\). The goal was to find the interval where \(| \frac{dx}{dt} | > | \frac{dy}{dt} |\). This inequality signifies that \(x\)'s rate of change is indeed faster than \(y\)'s within a certain interval.
By differentiating, we find the rates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), representing the speeds of change of \(x\) and \(y\) respectively. The relationship derived from this differentiation is given by \(3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}\). The goal was to find the interval where \(| \frac{dx}{dt} | > | \frac{dy}{dt} |\). This inequality signifies that \(x\)'s rate of change is indeed faster than \(y\)'s within a certain interval.
- Differentiation: Methods like differentiation are used to calculate rates of change, which reveal important relationships between variables.
- Applications: This technique is widely applied in various industries to predict trends and make informed decisions based on changing conditions.
Abscissa and Ordinate
In coordinate geometry, abscissa and ordinate are terms used to denote the coordinates of a point on the Cartesian plane. Abscissa refers to the horizontal coordinate \(x\), while ordinate refers to the vertical coordinate \(y\). These terms help in specifying the precise position of points within a plane by using an ordered pair \((x, y)\).
In our given exercise, the abscissa \(x\) changes at a faster rate than the ordinate \(y\). This scenario is common when analyzing movements or transformations where the horizontal activity is more rapid than the vertical.
In our given exercise, the abscissa \(x\) changes at a faster rate than the ordinate \(y\). This scenario is common when analyzing movements or transformations where the horizontal activity is more rapid than the vertical.
- Role in Equations: The terms abscissa and ordinate are crucial in formulating equations and analyzing graphs. They allow us to determine and predict the movement and positions of points.
- Practical Usage: Understanding these concepts is vital in fields like geography, engineering, and physics for mapping locations and designing trajectories.
Other exercises in this chapter
Problem 36
The minimum value of the function \(f(x)=\frac{x^{p}}{p}+\frac{x^{-q}}{q}\), where \(\frac{1}{p}+\frac{1}{q}=1, p>1\) is (A) 1 (B) 0 (C) 2 (D) None of these
View solution Problem 38
If a differentiable function \(f(x)\) has a relative minimum at \(x=0\), then the function \(y=f(x)+a x+b\) has a relative minimum at \(x=0\) for (A) all \(a>0\
View solution Problem 40
The maximum value of radius vector where \(\frac{c^{4}}{r^{2}}=\frac{a^{2}}{\sin ^{2} t}+\frac{b^{2}}{\cos ^{2} t} ;(a, b>0)\) is (A) \((a+b)^{2}\) (B) \(\frac{
View solution Problem 41
Let \(f(x)\) and \(g(x)\) be defined and differentiable for \(x \geq x_{0}\) and \(f\left(x_{0}\right)=g\left(x_{0}\right), f^{\prime}(x)>g^{\prime}(x)\) for \(
View solution