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(a) Find \(d^{2} y / d x^{2}\) for \(x=t^{3}+t, y=t^{2}\) (b) Is the curve concave up or down at \(t=1 ?\)

Short Answer

Expert verified
The second derivative is \( \frac{-6t^2 + 2}{(3t^2 + 1)^3}\). At \(t=1\), the curve is concave down.

Step by step solution

01

Find the Derivative of x with Respect to t

Given that \(x = t^3 + t\), find \(\frac{dx}{dt}\). The differentiation with respect to \(t\) gives us:\[\frac{dx}{dt} = 3t^2 + 1\]
02

Find the Derivative of y with Respect to t

Given that \(y = t^2\), find \(\frac{dy}{dt}\). The differentiation with respect to \(t\) gives us:\[\frac{dy}{dt} = 2t\]
03

Use Chain Rule to Find dy/dx

We use the chain rule to find \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\]Substitute \(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{3t^2 + 1}\):\[\frac{dy}{dx} = 2t \times \frac{1}{3t^2 + 1} = \frac{2t}{3t^2 + 1}\]
04

Differentiate dy/dx with Respect to t

Differentiate \(\frac{dy}{dx} = \frac{2t}{3t^2 + 1}\) with respect to \(t\) using the quotient rule:\[\frac{d}{dt}\left(\frac{2t}{3t^2 + 1}\right) = \frac{(3t^2 + 1)(2) - 2t(6t)}{(3t^2 + 1)^2}\]Simplify the numerator:\[= \frac{2(3t^2 + 1) - 12t^2}{(3t^2 + 1)^2} = \frac{-6t^2 + 2}{(3t^2 + 1)^2}\]
05

Find d^2y/dx^2 Using Chain Rule Again

Use the fact that \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}\):\[\frac{d^2y}{dx^2} = \left(\frac{-6t^2 + 2}{(3t^2 + 1)^2}\right) \times \frac{1}{3t^2+1}\]This simplifies to:\[\frac{d^2y}{dx^2} = \frac{-6t^2 + 2}{(3t^2 + 1)^3}\]
06

Evaluate d^2y/dx^2 at t=1

Substitute \(t = 1\) into \(\frac{d^2y}{dx^2}\) to check the concavity:\[\frac{d^2y}{dx^2} \bigg|_{t = 1} = \frac{-(6(1)^2) + 2}{(3(1)^2 + 1)^3} = \frac{-6 + 2}{4^3} = \frac{-4}{64} = -\frac{1}{16}\]
07

Determine Concavity

Since \(\frac{d^2y}{dx^2}\bigg|_{t = 1} = -\frac{1}{16}\) is negative, the curve is concave down at \(t = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative
In differential calculus, the second derivative, denoted as \( \frac{d^2y}{dx^2} \), provides valuable information about the behavior of a function's graph. While the first derivative \( \frac{dy}{dx} \) reveals the rate of change or slope of a function, the second derivative helps us understand how that slope is changing.

To calculate the second derivative for a parametric curve \( (x(t), y(t)) \), you need to first determine the first derivative \( \frac{dy}{dx} \) using the chain rule. Given functions \( x = x(t) \) and \( y = y(t) \), the chain rule tells us:
  • Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)
  • Calculate \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
Once you have \( \frac{dy}{dx} \), you differentiate it with respect to \( t \) to obtain \( \frac{d}{dt} \left( \frac{dy}{dx} \right) \). Finally, using the chain rule again, you multiply by \( \frac{dt}{dx} \) to find \( \frac{d^2y}{dx^2} \).

The second derivative helps assess the curvature of the graph. It's essential in determining areas such as concavity and points of inflection, providing deeper insights into the function's behavior.
Concavity
Concavity refers to the direction in which a curve opens. A graph can be concave up or concave down, and the second derivative helps us determine this property. Here's how concavity works:
  • A curve is concave up if \( \frac{d^2y}{dx^2} > 0 \). This means that the slope of the tangent to the curve is increasing as you move along the curve.
  • A curve is concave down if \( \frac{d^2y}{dx^2} < 0 \). Here, the slope of the tangent is decreasing.
Concavity essentially tells us about the acceleration of a function's rate of change. If \( \frac{d^2y}{dx^2} = 0 \), the function may have a point of inflection where the concavity changes. This transition could either be from concave up to concave down or vice versa.

In the original problem, when evaluated at \( t = 1 \), \( \frac{d^2y}{dx^2} = -\frac{1}{16} \). This negative value indicates that the curve is concave down at this point. Understanding concavity is crucial in fields like physics and economics, where the nature of a curve's bend dictates different interpretations.
Parametric Equations
Parametric equations are an elegant way to describe complex curves that can't easily be defined by a single function \( y = f(x) \). Instead, both \( x \) and \( y \) are given as functions of a third variable, typically \( t \), known as the parameter.

In our exercise, where \( x = t^3 + t \) and \( y = t^2 \), we see how parametric equations allow us to explore a curve in the \( xy \)-plane without directly linking \( x \) and \( y \). These equations are particularly useful to model motions and paths in physics or animations, where time \( t \) is a natural parameter.

To analyze a parametric curve's behavior, we often need to find derivatives, just like in regular Cartesian equations. Differentiation of parametric equations involves the chain rule, allowing us to link \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) to find the derivative \( \frac{dy}{dx} \). This step is crucial when examining points of tangency, concavity, or critical points on the curve.

Grasping the concept of parametric equations can be incredibly beneficial, as it equips students with the tools to tackle complex problems effectively, showcasing the power of mathematics in describing the natural world.

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Most popular questions from this chapter

Consider the family of functions \(y=a \cosh (x / a)\) for \(a>0 .\) Sketch graphs for \(a=1,2,3 .\) Describe in words the effect of increasing \(a\).

In a \(19^{\text {th }}\) century sea-battle, the number of ships on each side remaining \(t\) hours after the start are given by \(x(t)\) and \(y(t) .\) If the ships are equally equipped, the relation between them is \((x(t))^{2}-(y(t))^{2}=c,\) where \(c\) is a positive constant. The battle ends when one side has no ships remaining. (a) If, at the start of the battle, 50 ships on one side oppose 40 ships on the other, what is the value of \(c ?\) (b) If \(y(3)=16,\) what is \(x(3) ?\) What does this represent in terms of the battle? (c) There is a time \(T\) when \(y(T)=0 .\) What does this \(T\) represent in terms of the battle? (d) At the end of the battle, how many ships remain on the victorious side? (e) At any time during the battle, the rate per hour at which \(y\) loses ships is directly proportional to the number of \(x\) ships, with constant of proportionality k. Write an equation that represents this, Is \(k\) positive or negative? (f) Show that the rate per hour at which \(x\) loses ships is directly proportional to the number of \(y\) ships, with constant of proportionality \(k\) (g) Three hours after the start of the battle, \(x\) is losing ships at the rate of 32 ships per hour. What is \(k ?\) At what rate is \(y\) losing ships at this time?

Give an example of: A member of the family \(f(x)=a x^{3}-b x\) that has no critical points.

Find formulas for the functions described. A function of the form \(y=a \sin \left(b t^{2}\right)\) whose first critical point for positive \(t\) occurs at \(t=1\) and whose derivative is 3 when \(t=2\).

Grit, which is spread on roads in winter, is stored in mounds which are the shape of a cone. As grit is added to the top of a mound at 2 cubic meters per minute, the angle between the slant side of the cone and the vertical remains \(45^{\circ} .\) How fast is the height of the mound increasing when it is half a meter high? [Hint: Volume \(\left.V=\pi r^{2} h / 3, \text { where } r \text { is radius and } h \text { is height. }\right]\)

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