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Chapter 6: Problem 1

Sketch the graph of the given function. In each case determine whether \(f\) is continuous, piecewise continuous, or neither on the interval \(0 \leq t \leq 3 .\) $$ f(t)=\left\\{\begin{array}{ll}{t^{2},} & {0 \leq t \leq 1} \\ {2+t,} & {1

Short Answer

Expert verified
Given a function defined by: - \(f(t) = t^2\) for \(0 \leq t \leq 1\) - \(f(t) = 2+t\) for \(1 < t \leq 2\) - \(f(t) = 6-t\) for \(2 < t \leq 3\) After analyzing and sketching the function, we found that the function is discontinuous at \(t=1\) but continuous at \(t=2\). Therefore, the function is considered piecewise continuous on the interval \(0 \leq t \leq 3\).

Step by step solution

01

Sketch the graph of each piece of the function

We need to sketch the graph of the function in each of the given intervals: \(t^2\) for \(0 \leq t \leq 1\), \(2+t\) for \(1<t\leq 2\), and \(6-t\) for \(2<t\leq 3\).
02

Determine continuity at transition points

Now, we need to check the continuousness of the function at the transition points between the intervals. There are two points to check - \(t=1\) and \(t=2\). For \(t=1\), we have: - From the first interval: \(f(t) = t^2 \Rightarrow f(1) = 1^2 = 1\) - From the second interval: \(f(t) = 2 + t \Rightarrow f(1) = 2 + 1 = 3\) The function is discontinuous at \(t=1\) because the limits from the left and right do not match (\(1 \neq 3\)). For \(t=2\), we have: - From the second interval: \(f(t) = 2 + t \Rightarrow f(2) = 2 + 2 = 4\) - From the third interval: \(f(t) = 6 - t \Rightarrow f(2) = 6 - 2 = 4\) The function is continuous at \(t=2\) because the limits from the left and right match (\(4 = 4\)).
03

Determine the classification of the function on the interval \(0 \leq t \leq 3\)

Since the function is discontinuous at \(t=1\), it is not continuous on the interval \(0 \leq t \leq 3\). However, the function has only one point of discontinuity, so it is piecewise continuous on the interval \(0 \leq t \leq 3\).

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Most popular questions from this chapter

The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone-the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens \((1629-\) \(1695)\) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli's solution (in \(1690)\) was one of the first occasions on which a differential equation was explicitly solved. The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone-the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens \((1629-\) \(1695)\) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli's solution (in \(1690)\) was one of the first occasions on which a differential equation was explicitly solved. The geometrical configuration is shown in Figure \(6.6 .2 .\) The starting point \(P(a, b)\) is joined to the terminal point \((0,0)\) by the arc \(C .\) Arc length \(s\) is measured from the origin, and \(f(y)\) denotes the rate of change of \(s\) with respect to \(y:\) $$ f(y)=\frac{d s}{d y}=\left[1+\left(\frac{d x}{d y}\right)^{2}\right]^{1 / 2} $$ Then it follows from the principle of conservation of energy that the time \(T(b)\) required for a particle to slide from \(P\) to the origin is $$ T(b)=\frac{1}{\sqrt{2 g}} \int_{0}^{b} \frac{f(y)}{\sqrt{b-y}} d y $$

The Laplace transforms of certain functions can be found conveniently from their Taylor series expansions. (a) Using the Taylor series for \(\sin t\) $$ \sin t=\sum_{n=0}^{\infty} \frac{(-1)^{n} t^{2 n+1}}{(2 n+1) !} $$ and assuming that the Laplace transform of this scries can be computed term by term, verify that $$ \mathcal{L}\\{f(t)\\}=\arctan (1 / s), \quad s>1 $$ (c) The Bessel function of the first kind of order zero \(J_{0}\) has the Taylor series (see Section 5.8 ) $$ J_{0}(t)=\sum_{n=0}^{\infty} \frac{(-1)^{n} t^{2 n}}{2^{2 n}(n !)^{2}} $$ Assuming that the following Laplace transforms can be computed term by term, verify that $$ \mathcal{L}\left\\{J_{0}(t)\right\\}=\left(s^{2}+1\right)^{-1 / 2}, \quad s>1 $$ and $$\mathcal{L}\left[J_{0}(\sqrt{t})\right\\}=s^{-1} e^{-1 / 4 s}, \quad s>0$$

For each of the following initial value problems use the results of Problem 28 to find the differential equation satisfied by \(Y(s)=\mathcal{L}[\phi(t)\\},\) where \(y=\phi(t)\) is the solution of the given initial value problem. \(\begin{array}{ll}{\text { (a) } y^{\prime \prime}-t y=0 ;} & {y(0)=1, \quad y^{\prime}(0)=0 \text { (Airy's equation) }} \\ {\text { (b) }\left(1-t^{2}\right) y^{\prime \prime}-2 t y^{\prime}+\alpha(\alpha+1) y=0 ;} & {y(0)=0, \quad y^{\prime}(0)=1 \text { (Legendre's equation) }}\end{array}\) Note that the differential equation for \(Y(s)\) is of first order in part (a), but of second order in part (b). This is duc to the fact that \(t\) appcars at most to the first power in the equation of part (a), whereas it appears to the second power in that of part (b). This illustrates that the Laplace transform is not often useful in solving differential equations with variable coefficients, unless all the coefficients are at most linear functions of the independent variable.

Express the solution of the given initial value problem in terms of a convolution integral. \(y^{\prime \prime}+4 y^{\prime}+4 y=g(t) ; \quad y(0)=2, \quad y^{\prime}(0)=-3\)

Find the Laplace transform \(Y(s)=\mathcal{L}[y]\) of the solution of the given initial value problem. A method of determining the inverse transform is developed in Section \(6.3 .\) $$ y^{\prime \prime}+y=\left\\{\begin{array}{ll}{t,} & {0 \leq t<1,} \\ {0,} & {1 \leq t<\infty ; \quad y(0)=0, \quad y^{\prime}(0)=0}\end{array}\right. $$
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