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Tangent and concavity of parametric equations

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Intros
Lessons
  1. Tangent and Concavity of Parametric Equations Overview
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Examples
Lessons
  1. Find   dydx  \;\frac{dy}{dx}\; and   d2ydx2\;\frac{d^2y}{dx^2}
    1. x=t−t2x=t-t^2 , y=3+ty=3+t
    2. x=et x=e^t , y=e−ty=e^{-t}
  2. Questions Regarding to Tangents and Concavity
    Find the tangent to the cycloid x=r(θ−sin⁡θ)x=r(\theta - \sin \theta), y=r(1−cos⁡θ)y=r(1-\cos \theta) when θ=π4\theta = \frac{\pi}{4} and rr > 00. Determine the concavity for all values of θ\theta. (Do not eliminate the parameter)
    1. Find the point of the parametric curve x=t2+1x=t^2+1 and y=t3+t2y=t^3+t^2, in which the tangent is horizontal.
      1. Find the tangent to the curve x=3cos⁡tx=3 \cos t, y=4cos⁡ty=4 \cos t by:
        1. Eliminating the parameter.
        2. Without eliminating the parameter.
      Topic Notes
      ?
      In this lesson, we will focus on finding the tangent and concavity of parametric equations. Just like how we can take derivatives of Cartesian equations, we can also do it for parametric equations. First, we will learn to take the derivatives of parametric equations. Then we will look at an application which involves finding the tangents and concavity of a cycloid. After, we will look at special cases of finding a point with a horizontal tangent. Lastly, we will compare the difference of finding tangents by eliminating and without eliminating the parameter.

      Introduction

      Welcome to our exploration of tangent and concavity in parametric equations. This lesson begins with an essential introduction video that lays the foundation for understanding these complex concepts. Parametric equations offer a unique way to describe curves, and grasping their tangents and concavity is crucial for advanced calculus applications. We'll delve into the derivatives of parametric equations, a key tool for analyzing their behavior. You'll learn how to find tangents and determine the concavity of a cycloid, a fascinating curve with numerous real-world applications. We'll also examine special cases, particularly focusing on horizontal tangents, which provide valuable insights into the curve's shape and properties. By mastering these concepts, you'll enhance your ability to analyze and interpret parametric curves, opening doors to more advanced mathematical modeling and problem-solving techniques in various fields of study.

      Understanding the derivatives of parametric equations is essential for deeper insights into the behavior of these curves. By focusing on the analyzing parametric curves, you will be better equipped to tackle complex problems in calculus and beyond.

      Understanding Parametric Equations

      Parametric equations offer a unique way to describe curves and functions in mathematics, distinct from the more familiar Cartesian equations. Imagine you're on a journey, and instead of describing your location directly, you describe how far you've traveled in each direction as time passes. That's the essence of parametric equations!

      In Cartesian equations, we typically express y as a function of x, like y = f(x). However, parametric equations introduce a new variable, often denoted as 't', which acts as a parameter. Both x and y are then expressed as functions of this parameter: x = x(t) and y = y(t).

      Let's break this down with an example. Consider a simple circle. In Cartesian form, we might write it as x² + y² = r², where r is the radius. But parametrically, we could express it as x(t) = r cos(t) and y(t) = r sin(t), where t varies from 0 to 2π. As t changes, it traces out the circle.

      The beauty of parametric equations lies in their ability to describe more complex paths and motions. They're particularly useful in physics and engineering, where time-dependent motion is common. For instance, the path of a projectile can be elegantly described using parametric equations, with t representing time.

      One key advantage of parametric equations is their ability to represent curves that might be challenging or impossible to express as a single Cartesian equation. They can handle loops, self-intersections, and other complex shapes with ease. Think of a figure-eight or a spiral these are much simpler to describe parametrically.

      It's helpful to think of x(t) and y(t) as separate but related functions. As the parameter t changes, these functions work together to define points on the curve. This approach allows for a more dynamic and often more intuitive understanding of how curves are generated.

      In practice, you might encounter parametric equations in various forms. Sometimes, t represents time, as in motion problems. Other times, it might represent an angle, as in our circle example. The choice of parameter often depends on what makes the problem at hand easiest to solve or understand.

      Remember, while parametric equations might seem more complex at first, they often simplify problems and provide insights that Cartesian equations can't. As you explore more advanced mathematics and its applications, you'll find parametric equations to be a powerful tool in your mathematical toolkit.

      Derivatives of Parametric Equations

      Taking derivatives of parametric equations is an essential skill in calculus that allows us to analyze curves defined by two or more functions. In this section, we'll explore how to find the derivative dy/dx when x and y are given as functions of a parameter t. This process involves using the chain rule and a special formula that relates dy/dx to dy/dt and dx/dt.

      The Formula for dy/dx in Parametric Equations

      The key formula for finding the derivative of a parametric equation is:

      dy/dx = (dy/dt) / (dx/dt)

      This formula is derived from the chain rule and allows us to express dy/dx in terms of the derivatives with respect to the parameter t.

      Step-by-Step Instructions

      1. Identify the parametric equations for x(t) and y(t).
      2. Find dx/dt by differentiating x(t) with respect to t.
      3. Find dy/dt by differentiating y(t) with respect to t.
      4. Apply the formula: dy/dx = (dy/dt) / (dx/dt).
      5. Simplify the resulting expression if possible.

      Example

      Let's work through an example to illustrate this process:

      Given the parametric equations: x = t² + 1, y = t³ - t

      1. dx/dt = 2t (derivative of t²)
      2. dy/dt = 3t² - 1 (derivative of t³ - t)
      3. dy/dx = (3t² - 1) / (2t)
      4. Simplify: dy/dx = (3t² - 1) / 2t = (3t/2) - 1/(2t)

      Practice Problems

      To reinforce your understanding, try these practice problems:

      1. Find dy/dx if x = cos(t) and y = sin(t).
      2. Calculate dy/dx for x = e^t and y = ln(t).
      3. Determine dy/dx when x = t² - 2t and y = t³ + 3t².

      Applications of Parametric Derivatives

      Understanding how to take derivatives of parametric equations is crucial in various applications, including:

      • Analyzing the motion of objects in physics
      • Studying curves in computer graphics
      • Optimizing paths in engineering and robotics
      • Investigating economic models in finance

      Common Pitfalls and Tips

      When working with parametric derivatives, keep these points in mind:

      • Always check that dx/dt 0 before dividing.
      • Remember that t is the independent variable, not x or y.
      • Practice simplifying your final expressions for cleaner results.
      • Be prepared to use trigonometric identities when dealing with sine and cosine functions.

      Conclusion

      Mastering the technique of finding derivatives of parametric equations opens up a world of possibilities in calculus and its applications. By following the steps outlined above and practicing with various examples, you'll develop a strong foundation in this important concept. Remember, the key is to break down the process into manageable steps: find dx/dt and dy/dt separately, then combine them using the formula dy/dx = (dy/dt) / (dx/dt). With practice, you'll find that parametric equations practice problems become easier to solve.

      Finding Tangents of Parametric Curves

      Understanding how to find tangents of parametric curves is a crucial skill in calculus and analytical geometry. This process involves using the derivative formula and grasping the concepts of horizontal and vertical tangents. Let's explore this topic in depth and provide some practical examples and problems to solidify your understanding.

      The Derivative Formula for Parametric Curves

      When dealing with parametric curves, we have two functions x(t) and y(t) that define the curve. To find the tangent at any point, we use the formula:

      dy/dx = (dy/dt) / (dx/dt)

      This formula gives us the slope of the tangent line at any point on the curve. It's derived from the chain rule and is fundamental to analyzing parametric curves.

      Horizontal and Vertical Tangents

      Horizontal and vertical tangents are special cases that occur on parametric curves:

      • Horizontal Tangent: Occurs when dy/dt = 0 and dx/dt 0. The slope of the tangent line is zero.
      • Vertical Tangent: Occurs when dx/dt = 0 and dy/dt 0. The slope of the tangent line is undefined.

      Examples of Finding Tangents

      Example 1: Consider the parametric curve x = t^2, y = t^3. Find the slope of the tangent line at t = 1.

      Solution:

      1. dx/dt = 2t
      2. dy/dt = 3t^2
      3. dy/dx = (3t^2) / (2t) = 3t/2
      4. At t = 1, slope = 3/2

      Example 2: For the curve x = cos(t), y = sin(t), find where horizontal and vertical tangents occur.

      Solution:

      1. dx/dt = -sin(t)
      2. dy/dt = cos(t)
      3. Horizontal tangent: cos(t) = 0, occurs at t = π/2 and 3π/2
      4. Vertical tangent: sin(t) = 0, occurs at t = 0 and π

      Practice Problems

      To reinforce your understanding, try these problems:

      1. For the curve x = t^2 - 1, y = t^3 + 1, find the slope of the tangent line at t = 2.
      2. Determine the points where horizontal and vertical tangents occur for x = e^t, y = ln(t).
      3. Find the equation of the tangent line to the curve x = 3cos(t), y = 2sin(t) at t = π/4.

      Key Takeaways

      When working with tangents of parametric curves, remember:

      • Always use the derivative formula dy/dx = (dy/dt) / (dx/dt).
      • Check for horizontal tangents by setting dy/dt = 0.
      • Check for vertical tangents by setting dx/dt = 0.
      • Practice with various types of parametric equations to build proficiency.

      Mastering the technique of finding tangents to parametric curves is essential for advanced calculus and its applications in physics and engineering. By understanding the derivative formula and recognizing special cases like horizontal and vertical tangents

      Concavity of Parametric Curves

      Let's dive into the fascinating world of parametric curves and explore the concept of concavity! When we study parametric curves, we're looking at how they bend and curve in space. Concavity helps us understand the shape and behavior of these curves more deeply.

      So, what exactly is concavity? In simple terms, it's all about how a curve bends. A curve is concave up if it's shaped like a cup (), and concave down if it's shaped like an upside-down cup (). For parametric curves, determining concavity isn't as straightforward as with regular functions, but don't worry we've got a powerful tool to help us: the second derivative of parametric curves!

      To find the concavity of a parametric curve, we need to look at its second derivative of parametric curves with respect to x. Here's the formula:

      d²y/dx² = (y''t'² - y't''t') / (t')³

      Don't let this formula intimidate you! Let's break it down:

      • y'' and t'' are the second derivatives of y and t with respect to the parameter
      • y' and t' are the first derivatives of y and t with respect to the parameter
      • Remember, t' should not equal zero for this formula to work

      Now, how do we use this formula to determine concavity? It's actually quite simple:

      • If d²y/dx² > 0, the curve is concave up at that point
      • If d²y/dx² < 0, the curve is concave down at that point
      • If d²y/dx² = 0, we might have an inflection point (where the concavity changes)

      Let's look at an example to make this clearer. Suppose we have a parametric curve defined by x = t² and y = t³. To find the concavity:

      1. First, we calculate the derivatives:
        x' = 2t, x'' = 2
        y' = 3t², y'' = 6t
      2. Now, let's plug these into our formula:
        d²y/dx² = (6t(2t)² - 3t²(2)) / (2t)³
        = (24t³ - 6t²) / 8t³
        = (24 - 6/t) / 8
        = 3 - 3/(4t)
      3. Analyze the result:
        When t > 0, d²y/dx² > 0, so the curve is concave up
        When t < 0, d²y/dx² < 0, so the curve is concave down
        When t = 0, we have an inflection point

      Ready to practice? Try these parametric curve practice problems:

      1. Find the concavity of the curve x = t², y = t
      2. Determine any inflection points for x = t³, y = t²
      3. Sketch the curve x = cos(t), y = sin(t) and identify its concavity

      Remember, understanding concavity in parametric curves takes practice. Don't get discouraged if it seems tricky at first keep working through examples, and you'll soon develop an intuition for it. The key is to approach each problem step-by-step, calculate the derivatives carefully, and interpret the results. With time, you'll be able to visualize the concavity of parametric curves with ease!

      Application: Tangents and Concavity of a Cycloid

      Welcome to the fascinating world of cycloids! Today, we're going to explore how parametric equations come to life in a practical application. The cycloid, a curve traced by a point on the rim of a rolling wheel, is a perfect example of how parametric equations can describe real-world phenomena.

      Let's start by defining the parametric equations for a cycloid:

      x = a(t - sin t)
      y = a(1 - cos t)

      Here, 'a' represents the radius of the rolling circle, and 't' is our parameter. As t increases, we trace the path of the cycloid.

      Now, remember how we found tangent line slope for parametric curves? We can apply the same principle to our cycloid! To find the slope of the tangent line slope at any point, we use the formula:

      dy/dx = (dy/dt) / (dx/dt)

      For our cycloid, this gives us:

      dy/dx = (a sin t) / (a(1 - cos t)) = sin t / (1 - cos t)

      Isn't it amazing how our earlier methods seamlessly apply to this real-world curve?

      Let's move on to concavity. To determine the concavity of our cycloid, we need to find the second derivative. Remember, for parametric equations, we use:

      d²y/dx² = [(d²y/dt²)(dx/dt) - (dy/dt)(d²x/dt²)] / (dx/dt)³

      Applying this to our cycloid equations, we get:

      d²y/dx² = -2 / (a(1 - cos t))

      Now, let's put this knowledge into practice with a step-by-step example:

      Example: Find the slope of the tangent line slope and determine the concavity of the cycloid at t = π/3, given a = 2.

      Step 1: Calculate the slope
      dy/dx = sin(π/3) / (1 - cos(π/3)) = 3 / (1 - 1/2) = 23

      Step 2: Determine concavity
      d²y/dx² = -2 / (2(1 - cos(π/3))) = -2 / (2(1/2)) = -2

      Since the second derivative is negative, the cycloid is concave down at t = π/3.

      Great job following along! Now it's your turn to practice. Try these problems:

      1. Find the slope of the tangent line slope to the cycloid at t = π/4, given a = 1.
      2. Determine the concavity of the cycloid at t = π/2, given a = 3.
      3. At what values of t is the cycloid's tangent line slope horizontal?

      Remember, the key to mastering these concepts is practice. Don't be discouraged if you find it challenging at first that's all part of the learning process. Keep working through problems, and you'll soon find yourself comfortable with cycloids and their properties.

      As you explore cycloids further, you'll discover their fascinating properties and applications in physics and engineering. From the shape of roller coaster loops to the design of gear teeth, cycloids play a surprising role in our world. So keep up the great work you're not just solving math problems, you're unlocking the secrets of the curves all around us!

      Special Cases: Horizontal Tangents

      Horizontal tangents represent a special case in calculus that holds significant importance in various mathematical and real-world applications. These tangents occur when the slope of a curve at a particular point is zero, indicating a momentary "flatness" in the graph. Understanding horizontal tangents is crucial for analyzing function behavior and solving optimization problems.

      The key condition for identifying horizontal tangents is when the derivative of a function equals zero, expressed as dy/dt = 0. This condition signifies that the rate of change of y with respect to t is momentarily zero, resulting in a horizontal line tangent to the curve at that point. It's important to note that not all points where dy/dt = 0 are necessarily horizontal tangents; they could also represent vertical tangents or cusps.

      To illustrate this concept, consider the function f(x) = x³ - 3x² + 2. By setting its derivative f'(x) = 3x² - 6x to zero, we can find the x-coordinates of potential horizontal tangents. Solving 3x² - 6x = 0 yields x = 0 or x = 2, indicating horizontal tangents at these x-values. Graphically, these points represent local extrema of the function.

      Practice problems involving horizontal tangents often ask students to find points where the tangent line is horizontal or to determine the equations of these tangent lines. For example, given g(t) = t - 4t³ + 2, find the points where the tangent line is horizontal. To solve this, set g'(t) = 4t³ - 12t² = 0, factor out 4t², and solve for t to find the critical points.

      The significance of horizontal tangents extends beyond mathematical exercises. In physics, horizontal tangents can represent moments of zero velocity in projectile motion or turning points in oscillatory systems. In economics, they may indicate maximum profit or minimum cost points on production curves. Engineers use horizontal tangents to optimize designs, such as finding the most efficient shape for aerodynamic surfaces.

      Understanding horizontal tangents is also crucial in data analysis and machine learning. These points often represent local maxima or minima in datasets, helping identify trends, peaks, or valleys in complex data structures. In computer graphics and animation, horizontal tangents play a role in creating smooth curves and transitions, enhancing visual aesthetics and realism.

      As students delve deeper into calculus and its applications, mastering the concept of horizontal tangents becomes increasingly valuable. It forms the foundation for more advanced topics like optimization, inflection points, and curve sketching. By recognizing the condition dy/dt = 0 and its implications, learners gain a powerful tool for analyzing function behavior and solving real-world problems across various disciplines.

      Conclusion

      In this lesson, we've explored the fascinating world of parametric equations, focusing on tangents of parametric curves and concavity. We've learned how to analyze these equations, determine their behavior, and visualize their curves. Understanding tangents of parametric curves helps us grasp the direction and slope of parametric curves at specific points, while concavity reveals the shape and curvature of these functions. These concepts are crucial for various applications in mathematics and physics. To reinforce your understanding, we encourage you to revisit the introduction video for visual reinforcement and practice solving a variety of problems. Remember, mastering parametric equations requires consistent practice and application. As you continue your journey in mathematics, these skills will prove invaluable. We challenge you to explore more complex parametric equations and their real-world applications of parametric equations. Don't hesitate to seek additional resources or ask questions to deepen your understanding of this important topic.

      Tangent and Concavity of Parametric Equations

      Find   dydx  \;\frac{dy}{dx}\; and   d2ydx2\;\frac{d^2y}{dx^2} for the parametric equations x=t−t2x = t - t^2 and y=3+ty = 3 + t.

      Step 1: Find dydt\frac{dy}{dt} and dxdt\frac{dx}{dt}

      To find the first derivative dydx\frac{dy}{dx}, we need to determine dydt\frac{dy}{dt} and dxdt\frac{dx}{dt}.

      First, let's find dydt\frac{dy}{dt}. Given y=3+ty = 3 + t, the derivative of yy with respect to tt is:

      dydt=ddt(3+t)=0+1=1\frac{dy}{dt} = \frac{d}{dt}(3 + t) = 0 + 1 = 1

      Next, let's find dxdt\frac{dx}{dt}. Given x=t−t2x = t - t^2, the derivative of xx with respect to tt is:

      dxdt=ddt(t−t2)=1−2t\frac{dx}{dt} = \frac{d}{dt}(t - t^2) = 1 - 2t

      Step 2: Calculate dydx\frac{dy}{dx}

      Now that we have dydt\frac{dy}{dt} and dxdt\frac{dx}{dt}, we can find dydx\frac{dy}{dx} using the formula:

      dydx=dydtdxdt=11−2t\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{1 - 2t}

      So, the first derivative dydx\frac{dy}{dx} is 11−2t\frac{1}{1 - 2t}.

      Step 3: Find the Second Derivative d2ydx2\frac{d^2y}{dx^2}

      To find the second derivative d2ydx2\frac{d^2y}{dx^2}, we need to take the derivative of dydx\frac{dy}{dx} with respect to tt and then divide by dxdt\frac{dx}{dt}.

      First, let's find the derivative of dydx\frac{dy}{dx} with respect to tt. Given dydx=11−2t\frac{dy}{dx} = \frac{1}{1 - 2t}, we use the quotient rule:

      ddt(11−2t)=(1−2t)⋅0−1⋅(−2)(1−2t)2=2(1−2t)2\frac{d}{dt}\left(\frac{1}{1 - 2t}\right) = \frac{(1 - 2t) \cdot 0 - 1 \cdot (-2)}{(1 - 2t)^2} = \frac{2}{(1 - 2t)^2}

      Next, we divide this result by dxdt\frac{dx}{dt}:

      d2ydx2=ddt(dydx)dxdt=2(1−2t)21−2t\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{2}{(1 - 2t)^2}}{1 - 2t}

      To simplify, we multiply the numerator and the denominator by (1−2t)(1 - 2t):

      d2ydx2=2(1−2t)3\frac{d^2y}{dx^2} = \frac{2}{(1 - 2t)^3}

      So, the second derivative d2ydx2\frac{d^2y}{dx^2} is 2(1−2t)3\frac{2}{(1 - 2t)^3}.

      Summary

      In summary, we have found the first and second derivatives for the given parametric equations:

      First derivative: dydx=11−2t\frac{dy}{dx} = \frac{1}{1 - 2t}

      Second derivative: d2ydx2=2(1−2t)3\frac{d^2y}{dx^2} = \frac{2}{(1 - 2t)^3}

      FAQs

      Here are some frequently asked questions about tangents and concavity of parametric equations:

      1. What is the formula for finding the derivative of a parametric equation?

        The formula for finding the derivative of a parametric equation is dy/dx = (dy/dt) / (dx/dt). This formula allows us to find the slope of the tangent line at any point on a parametric curve.

      2. How do you determine the concavity of a parametric curve?

        To determine the concavity of a parametric curve, we use the second derivative formula: d²y/dx² = (y''t'² - y't''t') / (t')³. If d²y/dx² > 0, the curve is concave up; if d²y/dx² < 0, it's concave down; and if d²y/dx² = 0, it might be an inflection point.

      3. What are horizontal tangents in parametric equations?

        Horizontal tangents occur when dy/dt = 0 and dx/dt 0. At these points, the slope of the tangent line is zero, indicating a momentary "flatness" in the curve. These points often represent local maxima or minima of the function.

      4. How are parametric equations used in real-world applications?

        Parametric equations are used in various fields such as physics (describing projectile motion), engineering (designing curves for computer graphics), and economics (modeling complex relationships). They're particularly useful for describing paths that can't be easily represented by a single Cartesian equation.

      5. What is a cycloid and how is it related to parametric equations?

        A cycloid is a curve traced by a point on the rim of a rolling wheel. It's described by the parametric equations x = a(t - sin t) and y = a(1 - cos t), where 'a' is the radius of the wheel and 't' is the parameter. Cycloids are excellent examples of how parametric equations can model real-world phenomena.

      Prerequisite Topics

      Understanding the tangent and concavity of parametric equations is a crucial concept in advanced calculus, but it requires a solid foundation in several prerequisite topics. To fully grasp this subject, students must first familiarize themselves with key concepts that form the building blocks of this advanced topic.

      One of the fundamental prerequisites is defining curves with parametric equations. This concept is essential because it introduces the idea of representing curves using parameter-dependent functions. By understanding how to define curves parametrically, students can better visualize and analyze the behavior of these equations, including their tangents and concavity.

      The slope and equation of tangent line is another critical prerequisite. This topic provides the foundation for understanding how to determine the slope of a curve at any given point, which is crucial when analyzing the tangent of parametric equations. Mastering this concept allows students to transition smoothly into more complex applications involving parametric curves.

      A solid grasp of the chain rule is indispensable when dealing with parametric equations. The chain rule is a fundamental technique in calculus that allows for the differentiation of composite functions. In the context of parametric equations, it becomes especially important when calculating derivatives and determining the tangent lines of parametric curves.

      Additionally, familiarity with arc length and surface area of parametric equations provides valuable insights into the geometric properties of these curves. While this topic might seem more advanced, it helps in developing a comprehensive understanding of how parametric equations behave and how their properties relate to tangents and concavity.

      Understanding these prerequisite topics is crucial because they collectively build the framework necessary for analyzing the tangent and concavity of parametric equations. For instance, the ability to define curves parametrically allows students to represent complex shapes that might be difficult to express with standard functions. The knowledge of tangent lines and slopes directly applies to finding tangents of parametric curves, while the chain rule becomes an essential tool in differentiating these equations.

      Moreover, these prerequisites help in developing the analytical skills needed to interpret the behavior of parametric curves. The concept of concavity, which describes how a curve bends, becomes more intuitive when students have a strong foundation in these fundamental topics. By mastering these prerequisites, students can more easily visualize and understand the second derivative of parametric curves, which is key to determining concavity.

      In conclusion, a thorough understanding of these prerequisite topics is not just beneficial but essential for mastering the concept of tangent and concavity of parametric equations. Each topic contributes uniquely to building a comprehensive understanding, enabling students to tackle more complex problems and applications in advanced calculus with confidence and clarity.

      We can find the tangent (or derivative) without having to eliminate the parameter tt by using the equation:
      dydx=dydtdxdt   \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \; where   dxdt≠0\;\frac{dx}{dt} \neq0
      The horizontal tangent occurs when   dydt=0  \;\frac{dy}{dt} =0\; given that   dxdt≠0\;\frac{dx}{dt} \neq0.
      The vertical tangent occurs when   dxdt=0  \;\frac{dx}{dt} =0\; given that   dydt≠0\;\frac{dy}{dt} \neq0.

      To find the concavity (or second derivative), we use the following equation:
      d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}
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