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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. $$ \text{Parametric Equations} \quad \text{Parameter} $$ $$ x=t+1, y=t^{2}+3 t \quad t=-1 $$

Short Answer

Expert verified
The slope of the tangent line at \(t = -1\) in the parameter is 1, and the function is concave up at that point.

Step by step solution

01

Derive the Parametric Equations

Start by finding \(dy/dt\) and \(dx/dt\). From the given equations, \n\(dx/dt = 1\) \nand \n \(dy/dt = 2t + 3\).
02

Find dy/dx

The formula for \(dy/dx\) is \(\(dy/dx = dy/dt / dx/dt\)\). Therefore, plugging the derivatives into this equation yields: \n \(dy/dx = (2t + 3) / 1 = 2t + 3\).
03

Find second derivative

The formula for the second derivative \(d^2y/dx^2\) is \((d/dt(dy/dx))/dx/dt\). The derivative of \(dy/dx = 2t + 3\) is \(d(2t + 3)/dt = 2\). Therefore, \(d^2y/dx^2 = 2 / dx/dt = 2/1 = 2\).
04

Evaluate dy/dx and d^2y/dx^2 at t = -1

Substitute \(t = -1\) into \(dy/dx = 2t + 3\) to get the slope of the tangent at that point. This yields \(dy/dx = 2*(-1) + 3 = 1\). Also substitute \(t = -1\) into \(d^2y/dx^2 = 2\) to get the concavity at that point. Since \(d^2y/dx^2\) is a constant, its value does not change at \(t = -1\), so it also equals \(2\).
05

Identify slope and concavity

Since \(dy/dx = 1\) at \(t = -1\), the slope of the tangent line at that point is \(1\). Since \(d^2y/dx^2 = 2\) at \(t = -1\), the function is concave up at that point because \(d^2y/dx^2 > 0\).

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Key Concepts

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

Finding Slope

Finding the slope of a curve at a certain point is essential for understanding the behavior of the curve. The slope is the rate at which the curve rises or falls as you move along it. In calculus, to find the slope of a curve defined by parametric equations, such as x = t + 1, y = t^2 + 3t, we use the derivative of y with respect to x, denoted as dy/dx. This derivative is calculated by first finding dy/dt and dx/dt, and then using the formula dy/dx = dy/dt / dx/dt.

  • To obtain the slope at a specified parameter value, simply substitute that value into the dy/dx expression.

For instance, the solution above demonstrates that at parameter t = -1, the slope is 1, suggesting a relatively steady increase of the curve at that point.

Concavity

Concavity refers to the direction in which a curve bends. Mathematically, it is determined by the second derivative, specifically d^2y/dx^2. If the second derivative is positive, the curve is said to be concave up, resembling a U-shaped curve. On the other hand, if it is negative, the curve is concave down, like an upside-down U.

  • A zero second derivative indicates that the curve may shift from concave up to concave down or vice versa, or it could be a point of inflection where the curve changes its bending orientation.

The example in the solution illustrates a constant second derivative of 2, which means the curve is consistently concave up at t = -1, representing a U-shape around that parameter value.

Second Derivative

The second derivative of a function provides insights into its curvature and concavity. In the context of parametric equations, the second derivative d^2y/dx^2 is obtained by differentiating dy/dx with respect to t, and then dividing by dx/dt. The second derivative can reveal points of inflection where the concavity changes.

  • For practical use, when we find a constant second derivative as in the solution provided, such as 2, we know the curve does not have points of inflection and maintains the same concavity throughout.

This continuity in concavity is critical for understanding the overall shape and behavior of the graph of the function.

Tangent Line

The tangent line to a curve at a specific point is a straight line that just touches the curve at that point and has the same slope as the curve at that point. It provides an approximation to the curve near that point and is crucial for finding instantaneous rates of change. The slope of the tangent line at any point is found by substituting the parameter value into the derivative dy/dx.

  • If the parameter value changes, we can find a new tangent slope to give us different instantaneous rates of changes at different points of the curve.

In the solution, we've determined that at t = -1, the slope of the tangent line is 1, which means that the tangent is inclined at a 45-degree angle, rising as it moves to the right.

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Most popular questions from this chapter

Proof (a) Prove that if any two tangent lines to a parabola intersect at right angles, their point of intersection must lie on the directrix. (b) Demonstrate the result of part (a) by showing that the tangent lines to the parabola \(x^{2}-4 x-4 y+8=0\) at the points \((-2,5)\) and \(\left(3, \frac{5}{4}\right)\) intersect at right angles, and that the point of intersection lies on the directrix.

Finding Equations of Tangent Lines In Exercises \(15-18\) , find an equation of the tangent line at each given point on the curve. $$ \begin{array}{l}{x=2-3 \cos \theta, \quad y=3+2 \sin \theta} \\\ {(-1,3),(2,5),\left(\frac{4+3 \sqrt{3}}{2}, 2\right)}\end{array} $$

In Exercises 75–77, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The curve represented by the parametric equations \(x=t\) and \(y=\cos t\) can be written as an equation of the form \(y=f(x)\)

The radiation from a transmitting antenna is not uniform in all directions. The intensity from a particular antenna is modeled by \(r=a \cos ^{2} \theta\) (a) Convert the polar equation to rectangular form. (b) Use a graphing utility to graph the model for \(a=4\) and \(a=6\) . (c) Find the area of the geographical region between the two curves in part (b).

Finding the Arc Length of a Polar Curve In Exercises \(51-56,\) find the length of the curve over the given interval. $$ \begin{array}{ll}{\text { Polar Equation }} & {\text { Interval }} \\ {r=4 \sin \theta} & {0 \leq \theta \leq \pi}\end{array} $$

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