Problem 8 Finding Slope and Concavity In E... [FREE SOLUTION]

91影视

Calculus of a Single Variable

Ron Larson, Bruce Edwards

$Math Studyset 91影视 Explanations$ Math

10 Edition

Chapter 10: Problem 8

Finding Slope and Concavity In Exercises $5-14$ , find $d y / d x$ and $d^{2} y / d x^{2},$ and find the slope and concavity (if possible) at the given value of the parameter. $$ \text{Parametric Equations} \quad \text{Parameter} $$ $$ x=t^{2}+5 t+4, y=4 t \quad t=0 $$

Short Answer

Expert verified

At $t = 0$, the slope of $y$ with respect to $x$ is $4/5$ which indicates an increasing slope, and the concavity is $-4/25$ which indicates a concave down.

Step by step solution

Find dx/dt and dy/dt

First, the derivatives of $x$ and $y$ with respect to the parameter $t$ need to be computed. Using the given equations $x=t^{2}+5t+4$ and $y=4t$, we can find $dx/dt = 2t+5$ and $dy/dt = 4$.

Find dy/dx

Next, compute $dy/dx$ using the formula $dy/dx = (dy/dt) / (dx/dt)$. Using the results from Step 1, we obtain $dy/dx = 4 / (2t+5)$.

Find d^2y/dx^2

For the second derivative, it's important to be aware that $d^2y/dx^2$ is not the derivative of $dy/dx$ with respect to $t$. Instead, you use the formula $d^2y/dx^2 = (d/dt(dy/dx)) / (dx/dt)$. Before that, calculate the derivative of $dy/dx$ with respect to $t$ and we get $-4/(2t+5)^2$. Substituting $dx/dt$ into the formula, we get $d^2y/dx^2 = (-4/(2t+5)^2) / (2t+5)$.

Find the slope and concavity at t=0

Now, obtain the slope and the concavity at the given value of $t=0$ by substituting $t=0$ into $dy/dx$ and $d^2y/dx^2$ respectively. The results are $dy/dx = 4 / 5$ and $d^2y/dx^2 = -4 / 25$. The positive value of $dy/dx$ indicates an increasing slope while the negative value of $d^2y/dx^2$ indicates a concave down.

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Key Concepts

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

Differentiation

Differentiation is a fundamental concept in calculus focusing on how functions change. It deals with finding the derivative, which represents the rate of change of a function with respect to a variable. When dealing with parametric equations such as

$x = t^2 + 5t + 4$
$y = 4t$

,it's crucial to consider the parameter, in this case, $t$, to determine how the changes in $x$ and $y$ relate to each other. Differentiation helps us find

$\frac{dx}{dt}$
$\frac{dy}{dt}$

,allowing us to compute the slope and concavity at a particular parameter value.

Slope

The slope represents how steep a line is at a given point. It is a measure of the function's rate of change with respect to a variable. When dealing with parametric equations, the slope $\frac{dy}{dx}$ is determined by the ratio $\frac{dy/dt}{dx/dt}$. From our parametric equations, this becomes

$\frac{dy}{dx} = \frac{4}{2t + 5}$

The slope helps in understanding whether a function is increasing (positive slope) or decreasing (negative slope) at a specific point. For $t = 0$, the slope is $\frac{4}{5}$, indicating a mild increase.

Concavity

Concavity indicates the direction and curvature of a function. It helps in understanding whether the graph of a function is bending upwards or downwards. The second derivative $\frac{d^2y}{dx^2}$ enlightens us about the function鈥檚 concavity. A positive second derivative suggests that the function is concave up (bending upwards), while a negative one indicates concave down (bending downwards).

First Derivative

The first derivative of a function provides critical information about the slope. In our case, we use parametric equations and differentiate each part with respect to the parameter $t$. From

$x = t^2 + 5t + 4$
$y = 4t$

,we find the derivatives:

$\frac{dx}{dt} = 2t + 5$
$\frac{dy}{dt} = 4$

. Hence,

$\frac{dy}{dx} = \frac{4}{2t + 5}$

. At $t = 0$, this evaluates to $\frac{4}{5}$. It tells us that at this point, the function is increasing, as indicated by the positive value.

Second Derivative

The second derivative is a deeper exploration into how the function's rate of change is itself changing. It's like a derivative of a derivative. For parametric equations, the formula

$\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$

is applied. With our given

$dy/dx = 4 / (2t + 5)$

, we first find

$d/dt(dy/dx) = - 4 / (2t + 5)^2$

,then use

$\frac{d^2y}{dx^2} = \frac{- 4 / (2t + 5)^2}{2t + 5}$

, which at $t = 0$ results in $-4/25$. This negative value indicates the curve at the point $t=0$ is concave down. Such insight is invaluable in sketching graphs and predicting behavior trends.

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91影视

Finding Slope and Concavity In Exercises \(5-14\) , find \(d y / d x\) and \(d^{2} y / d x^{2},\) and find the slope and concavity (if possible) at the given value of the parameter. $$ \text{Parametric Equations} \quad \text{Parameter} $$ $$ x=t^{2}+5 t+4, y=4 t \quad t=0 $$

Short Answer

Step by step solution

Find dx/dt and dy/dt

Find dy/dx

Find d^2y/dx^2

Find the slope and concavity at t=0

Key Concepts

Differentiation

Slope

Concavity

First Derivative

Second Derivative

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