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Chapter 11: Problem 6

Determine the interval on which \(y=t^{3}\) is (a) concave up, (b) concave down.

Short Answer

Expert verified
(a) Concave up on \((0, \infty)\). (b) Concave down on \((-\infty, 0)\).

Step by step solution

01

Find the second derivative

To determine concavity, we first need to calculate the second derivative of the function. The given function is \[ y = t^3 \] The first derivative of \( y \) with respect to \( t \) is: \[ y' = \frac{d}{dt}(t^3) = 3t^2 \] The second derivative of \( y \) is: \[ y'' = \frac{d}{dt}(3t^2) = 6t \]
02

Determine when the second derivative is positive or negative

Concave up intervals are where the second derivative \( y'' \) is positive, and concave down intervals are where \( y'' \) is negative. The second derivative is \[ y'' = 6t \] Set this greater than 0 for concave up: \[ 6t > 0 \]Divide both sides by 6, we get \[ t > 0 \] Set it less than 0 for concave down: \[ 6t < 0 \]Divide both sides by 6, we get \[ t < 0 \]
03

State the intervals of concavity

The function \( y = t^3 \) is concave up on the interval where \( t > 0 \), which is \((0, \infty)\).The function is concave down on the interval where \( t < 0 \), which is \((-\infty, 0)\).

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Key Concepts

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

Second Derivative
The second derivative is a powerful tool in calculus, used to determine the concavity of a function. It is obtained by differentiating the first derivative of a function. For the function given, \( y = t^3 \), the first step is finding the first derivative, which is \( y' = 3t^2 \). Differentiating once more, we arrive at the second derivative: \( y'' = 6t \). This expression will help us understand how the graph of the function curves. Essentially, the sign of the second derivative (positive or negative) at different points tells us whether the function is concave up or concave down in those intervals.
First Derivative
Before diving into concavity, we must first understand the first derivative of a function. The first derivative represents the rate of change of the function, or the slope of the tangent line at any point on the curve. For \( y = t^3 \), the first derivative is \( y' = 3t^2 \). This derivative shows us how steep the curve is at any given point. However, for concavity, the first derivative alone isn't enough. We need to evaluate the second derivative to make specific claims about the curve's bending directions.
Concave Up
A function is said to be concave up on an interval if its graph bends upwards like a cup, and the second derivative is positive in that interval. For the equation \( y'' = 6t \) that we derived, we check where it is positive. Setting \( 6t > 0 \), we find that \( t > 0 \). This indicates that for \( t > 0 \) (specifically, on the interval \((0, \infty)\)), the graph of \( y = t^3 \) is concave up. The curve lifts upward, suggesting that the rate of change is increasing.
Concave Down
On the flip side, a function is concave down on an interval if its graph bends downward like an umbrella, indicated by a negative second derivative. Again using \( y'' = 6t \), we explore where \( 6t < 0 \), resulting in \( t < 0 \). This shows that for \( t < 0 \) (on the interval \((-\infty, 0)\)), the function \( y = t^3 \) is concave down. Here, the graph bends downwards, signifying that the rate of change is decreasing in this section.

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Most popular questions from this chapter

\text { If } x=t+t^{2}+t^{3} \text { and } y=\sin 2 t \text {, find } \frac{\mathrm{d} y}{\mathrm{~d} x} \text { when } t=1 \text {. }

Differentiate (a) \(y=\ln x\) (b) \(y=\ln 2 x\) (c) \(y=\ln k x, k\) constant (d) \(y=\ln (1+t)\) (e) \(y=\ln (3+4 t)\) (f) \(y=\ln (5+7 \sin x)\)

If $$ y=2 x^{3}+3 x^{2}-12 x+1 $$ find values of \(x\) for which \(y^{\prime \prime}=0\)

Use logarithmic differentiation to find the derivatives of the following functions: (a) \(y=x^{4} \mathrm{e}^{x}\) (b) \(y=\frac{1}{x} \mathrm{e}^{-x}\) (c) \(z=t^{3}(1+t)^{9}\) (d) \(y=\mathrm{e}^{x} \sin x\) (e) \(y=x^{7} \sin ^{4} x\)

Use logarithmic differentiation to find the derivatives of the following functions: (a) \(z=t^{4}(1-t)^{6}(2+t)^{4}\) (b) \(y=\frac{\left(1+x^{2}\right)^{3} \mathrm{e}^{7 x}}{(2+x)^{6}}\) (c) \(x=(1+t)^{3}(2+t)^{4}(3+t)^{5}\) (d) \(y=\frac{\left(\sin ^{4} t\right)\left(2-t^{2}\right)^{4}}{\left(1+e^{t}\right)^{6}}\) (e) \(y=x^{3} e^{x} \sin x\)
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