?Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=1+y\) Equation Transcription: Text Transcription: 1-10 y^prime = 1 + y

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=y-y^{3}\) Equation Transcription: Text Transcription: 1-10 y^prime = y-y^3

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=\sin (t+y)\) Equation Transcription: Text Transcription: 1-10 y^prime = sin (t+y)

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=\ln (t y)\) Equation Transcription: Text Transcription: 1-10 y^prime = ln(ty)

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=e^{t+y}\) Equation Transcription: Text Transcription: 1-10 y^prime = e^t+y

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=\frac{y+1}{y}+y\) Equation Transcription: Text Transcription: 1-10 y^prime = y+1 / ty + y

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=\frac{e^{t+y}}{y+1}\) Equation Transcription: Text Transcription: 1-10 y^prime = e^t+y / y+1

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=t \ln \left(y^{2 t}\right)+t^{2}\) Equation Transcription: Text Transcription: 1-10 y^prime = t ln (y^2t) + t^2

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(y^{\prime}=\frac{y}{t}+\frac{t}{y}\) Equation Transcription: Text Transcription: 1-10 y^prime = y/t + t/y

Separable or Not Determine whether each equation in Problems \(1-10\) is separable or not. Write separable ones in separable form. Determine constant solutions, if any. \(t y^{\prime}=1+y^{2}\) Equation Transcription: Text Transcription: 1-10 t y^prime = 1 + y^2

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(y^{\prime}=\frac{t^{2}}{y}\) Equation Transcription: Text Transcription: DE IVP 11-20 y y^prime = t^2 / y

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(t y^{\prime}=\sqrt{1-y^{2}}\) Equation Transcription: Text Transcription: DE IVP 11-20 y t y^prime = sqrt 1 - y^2

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(y^{\prime}=\frac{x^{2}+7}{y^{4}-4 y^{3}}\) Equation Transcription: Text Transcription: DE IVP 11-20 y y^prime = t^2 + 7 / y^4 - 4y^3

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(t y^{\prime}=4 y\) Equation Transcription: Text Transcription: DE IVP 11-20 y t y^prime = 4y

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(y^{\prime}=y \cos t\) Equation Transcription: Text Transcription: DE IVP 11-20 y y^prime = y cos t

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(4 t d y=\left(y^{2}+t y^{2}\right) d t, \quad y(1)=1\) Equation Transcription: Text Transcription: DE IVP 11-20 y 4tdy = (y^2 + ty^2)dt, y(1)=1

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(y^{\prime}=\frac{1-2 t}{y}, \quad v(1)=-2\) Equation Transcription: Text Transcription: DE IVP 11-20 y y^prime = 1-2t / y, v(1) = -2

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(y^{\prime}=y^{2}-4, \quad v(0)=0\) Equation Transcription: Text Transcription: DE IVP 11-20 y y^prime = y^2 - 4, v(0)=0

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(y^{\prime}=\frac{2 t}{1+2 y}, \quad y(2)=0\) Equation Transcription: Text Transcription: DE IVP 11-20 y y^prime = 2t / 1+2y, y(2) = 0

Solving by Separation Use separation of variables to obtain solutions to the \(DE\)s and \(IVP\)s in Problems \(11-20\). Solve for \(y\) when possible. \(y^{\prime}=\frac{1+y^{2}}{1+t^{2}}, \quad y(0)=-1\) Equation Transcription: Text Transcription: DE IVP 11-20 y y^prime = 1+y^2 / 1+t^2, y(0) = -1

Integration by Parts Problems \(21-24\) involve a variety of integration techniques. Recall that the formula for integration by parts is \(\int u d v=u v-\int v d u\), where the variables \(u\) and \(dv\) must be assigned carefully. Determine the solutions to the following \(DE\)s. \(y^{\prime}=\left(\cos ^{2} y\right) \ln t\) Equation Transcription: Text Transcription: 21-24 integral u dv = uv - integral v du u dv DE y^prime = (cos^2 y) ln t

Integration by Parts Problems \(21-24\) involve a variety of integration techniques. Recall that the formula for integration by parts is \(\int u d v=u v-\int v d u\), where the variables \(u\) and \(dv\) must be assigned carefully. Determine the solutions to the following \(DE\)s. \(y^{\prime}=\left(t^{2}-5\right) \cos 2 t\) Equation Transcription: Text Transcription: 21-24 integral u dv = uv - integral v du u dv DE y^prime = (t^2 - 5) cos 2t

Integration by Parts Problems \(21-24\) involve a variety of integration techniques. Recall that the formula for integration by parts is \(\int u d v=u v-\int v d u\), where the variables \(u\) and \(dv\) must be assigned carefully. Determine the solutions to the following \(DE\)s. \(y^{\prime}=t^{2} e^{v+2 t}\) Equation Transcription: Text Transcription: 21-24 integral u dv = uv - integral v du u dv DE y^prime = t^2 e^v+2t

Integration by Parts Problems \(21-24\) involve a variety of integration techniques. Recall that the formula for integration by parts is \(\int u d v=u v-\int v d u\), where the variables \(u\) and \(dv\) must be assigned carefully. Determine the solutions to the following \(DE\)s. \(y^{\prime}=t v e^{-t}\) Equation Transcription: Text Transcription: 21-24 integral u dv = uv - integral v du u dv DE y^prime = tve^-t

Equilibria and Direction Fields Match each autonomous \(DE\) in Problems \(25-30\) with a direction field in Figure \(1.3.6\). HINT: First determine the equilibrium solutions and then determine the signs of slopes for appropriate values of \(y\). \(y^{\prime}=1-y^{2}\) Equation Transcription: Text Transcription: DE 25-30 1.3.6 y y^prime=1-y^2

Equilibria and Direction Fields Match each autonomous \(DE\) in Problems \(25-30\) with a direction field in Figure \(1.3.6\). HINT: First determine the equilibrium solutions and then determine the signs of slopes for appropriate values of \(y\). \(y^{\prime}=y^{2}-1\) Equation Transcription: Text Transcription: DE 25-30 1.3.6 y y^prime=y^2-1

Equilibria and Direction Fields Match each autonomous \(DE\) in Problems \(25-30\) with a direction field in Figure \(1.3.6\). HINT: First determine the equilibrium solutions and then determine the signs of slopes for appropriate values of \(y\). \(y^{\prime}=y(y-1)(y+1)\) Equation Transcription: Text Transcription: DE 25-30 1.3.6 y y^prime=y(y-1)(y+1)

Equilibria and Direction Fields Match each autonomous \(DE\) in Problems \(25-30\) with a direction field in Figure \(1.3.6\). HINT: First determine the equilibrium solutions and then determine the signs of slopes for appropriate values of \(y\). \(y^{\prime}=(y-1)^{2}\) Equation Transcription: Text Transcription: DE 25-30 1.3.6 y y^prime=(y-1)^2

Equilibria and Direction Fields Match each autonomous \(DE\) in Problems \(25-30\) with a direction field in Figure \(1.3.6\). HINT: First determine the equilibrium solutions and then determine the signs of slopes for appropriate values of \(y\). \(y^{\prime}=(y+1)(y-1)^{2}\) Equation Transcription: Text Transcription: DE 25-30 1.3.6 y y^prime=(y+1)(y-1)^2

Equilibria and Direction Fields Match each autonomous \(DE\) in Problems \(25-30\) with a direction field in Figure \(1.3.6\). HINT: First determine the equilibrium solutions and then determine the signs of slopes for appropriate values of \(y\). \(y^{\prime}=\left(y^{2}+1\right)(y-1)\) Equation Transcription: Text Transcription: DE 25-30 1.3.6 y y^prime=(y^2 + 1)(y-1)

Finding the Nonequilibrium Solutions Solve the \(DE\)s in Problems \(31-34\). Most of these solutions will require the use of the method of partial fractions (See Appendix PF for a review of this material.) Write down the equilibrium solutions as well as the nonequilibrium solutions for each \(DE\). NOTE: For most of these equations, direction fields are drawn with the previous group of problems. \(y^{\prime}=1-y^{2}\) Equation Transcription: Text Transcription: DE 31-34 y^prime = 1-y^2

Finding the Nonequilibrium Solutions Solve the \(DE\)s in Problems \(31-34\). Most of these solutions will require the use of the method of partial fractions (See Appendix PF for a review of this material.) Write down the equilibrium solutions as well as the nonequilibrium solutions for each \(DE\). NOTE: For most of these equations, direction fields are drawn with the previous group of problems. \(y^{\prime}=2 y-y^{2}\) Equation Transcription: Text Transcription: DE 31-34 y^prime = 2y-y^2

Finding the Nonequilibrium Solutions Solve the \(DE\)s in Problems \(31-34\). Most of these solutions will require the use of the method of partial fractions (See Appendix PF for a review of this material.) Write down the equilibrium solutions as well as the nonequilibrium solutions for each \(DE\). NOTE: For most of these equations, direction fields are drawn with the previous group of problems. \(y^prime = y(y-1)(y+1)\) Equation Transcription: Text Transcription: DE 31-34 y^prime = y(y-1)(y+1)

Finding the Nonequilibrium Solutions Solve the \(DE\)s in Problems \(31-34\). Most of these solutions will require the use of the method of partial fractions (See Appendix PF for a review of this material.) Write down the equilibrium solutions as well as the nonequilibrium solutions for each \(DE\). NOTE: For most of these equations, direction fields are drawn with the previous group of problems. \(y^{\prime}=(y-1)^{2}\) Equation Transcription: Text Transcription: DE 31-34 y^prime = (y-1)^2

Help from Technology For each \(DE\) in Problems \(35-40\), solve analytically to obtain solution curves through the points \((1,1)\) and \((-1,-1)\). Then, using an appropriate software package, draw the direction field and superimpose your solution curves onto it. \(\frac{d y}{d t}=y\) Equation Transcription: Text Transcription: DE 35-40 (1,1) (-1,-1) dy/dt=y

Help from Technology For each \(DE\) in Problems \(35-40\), solve analytically to obtain solution curves through the points \((1,1)\) and \((-1,-1)\). Then, using an appropriate software package, draw the direction field and superimpose your solution curves onto it. \(\frac{d y}{d t}=\cos t\) Equation Transcription: Text Transcription: DE 35-40 (1,1) (-1,-1) dy/dt= cos t

Help from Technology For each \(DE\) in Problems \(35-40\), solve analytically to obtain solution curves through the points \((1,1)\) and \((-1,-1)\). Then, using an appropriate software package, draw the direction field and superimpose your solution curves onto it. \(\frac{d y}{d t}=\frac{t}{y^{2} \sqrt{1+t^{2}}}\) Equation Transcription: Text Transcription: DE 35-40 (1,1) (-1,-1) dy/dt= t/ y^2 sqrt 1+t^2

Help from Technology For each \(DE\) in Problems \(35-40\), solve analytically to obtain solution curves through the points \((1,1)\) and \((-1,-1)\). Then, using an appropriate software package, draw the direction field and superimpose your solution curves onto it. \(\frac{d y}{d t}=y \cos t\) Equation Transcription: Text Transcription: DE 35-40 (1,1) (-1,-1) dy/dt= y cos t

Help from Technology For each \(DE\) in Problems \(35-40\), solve analytically to obtain solution curves through the points \((1,1)\) and \((-1,-1)\). Then, using an appropriate software package, draw the direction field and superimpose your solution curves onto it. \(\frac{d y}{d t}=\frac{2 t(y+1)}{y}\) Equation Transcription: Text Transcription: DE 35-40 (1,1) (-1,-1) dy/dt= 2t(y+1) / y

Help from Technology For each \(DE\) in Problems \(35-40\), solve analytically to obtain solution curves through the points \((1,1)\) and \((-1,-1)\). Then, using an appropriate software package, draw the direction field and superimpose your solution curves onto it. \(\frac{d y}{d t}=\sin (t y)\) Equation Transcription: Text Transcription: DE 35-40 (1,1) (-1,-1) dy/dt= sin (ty)

Making Equations Separable Many differential equations that are not separable can be made separable by making a proper substitution. One example is the class of first-order equations with right-hand sides that are functions of the combination \(y/t\) (or \(t/y\)). Given such a \(DE\) \(\frac{d y}{d t}=f\left(\frac{y}{t}\right)\), called Euler-homogenous, let \(v=y / t\). By the product rule, we deduce from \(\mathrm{y}=\mathrm{vt}\) that \(\frac{d y}{d t}=v+t \frac{d v}{d t}\), so the equation becomes \(v+t \frac{d v}{d t}=f(v)\), which separates into \(\frac{d t}{t}=\frac{d v}{f(v)-v}\). Apply this method to solve the Euler-homogeneous \(DE\)s and \(IVP\)s in Problems \(41-44\). Plot sample solutions on a direction field and discuss. \(\frac{d y}{d t}=\frac{y+t}{t}\) Equation Transcription: Text Transcription: y/t t/y dy/dt=f(y/t) v=y/t y=vt dy/dt=v+t dv/dt v+t dv/dt=f (v) d t/t=d v / f (v) -v DE IVP 41-44 dy/dt = y+t / t

Making Equations Separable Many differential equations that are not separable can be made separable by making a proper substitution. One example is the class of first-order equations with right-hand sides that are functions of the combination \(y/t\) (or \(t/y\)). Given such a \(DE\) \(\frac{d y}{d t}=f\left(\frac{y}{t}\right)\), called Euler-homogenous, let \(v=y / t\). By the product rule, we deduce from \(\mathrm{y}=\mathrm{vt}\) that \(\frac{d y}{d t}=v+t \frac{d v}{d t}\), so the equation becomes \(v+t \frac{d v}{d t}=f(v)\), which separates into \(\frac{d t}{t}=\frac{d v}{f(v)-v}\). Apply this method to solve the Euler-homogeneous \(DE\)s and \(IVP\)s in Problems \(41-44\). Plot sample solutions on a direction field and discuss. \(\frac{d y}{d t}=\frac{y^{2}+t^{2}}{y t}, \quad y(1)=-2\) Equation Transcription: Text Transcription: y/t t/y dy/dt=f(y/t) v=y/t y=vt dy/dt=v+t dv/dt v+t dv/dt=f (v) d t/t=d v / f (v) -v DE IVP 41-44 dy/dt=y^2+t^2/yt, y(1) = -2

Making Equations Separable Many differential equations that are not separable can be made separable by making a proper substitution. One example is the class of first-order equations with right-hand sides that are functions of the combination \(y/t\) (or \(t/y\)). Given such a \(DE\) \(\frac{d y}{d t}=f\left(\frac{y}{t}\right)\), called Euler-homogenous, let \(v=y / t\). By the product rule, we deduce from \(\mathrm{y}=\mathrm{vt}\) that \(\frac{d y}{d t}=v+t \frac{d v}{d t}\), so the equation becomes \(v+t \frac{d v}{d t}=f(v)\), which separates into \(\frac{d t}{t}=\frac{d v}{f(v)-v}\). Apply this method to solve the Euler-homogeneous \(DE\)s and \(IVP\)s in Problems \(41-44\). Plot sample solutions on a direction field and discuss. \(\frac{d y}{d t}=\frac{2 v^{4}+t^{4}}{v^{3}}\) Equation Transcription: Text Transcription: y/t t/y dy/dt=f(y/t) v=y/t y=vt dy/dt=v+t dv/dt v+t dv/dt=f (v) d t/t=d v / f (v) -v DE IVP 41-44 dy/dt=2y^4 + t^4 /ty^3

Making Equations Separable Many differential equations that are not separable can be made separable by making a proper substitution. One example is the class of first-order equations with right-hand sides that are functions of the combination \(y/t\) (or \(t/y\)). Given such a \(DE\) \(\frac{d y}{d t}=f\left(\frac{y}{t}\right)\), called Euler-homogenous, let \(v=y / t\). By the product rule, we deduce from \(\mathrm{y}=\mathrm{vt}\) that \(\frac{d y}{d t}=v+t \frac{d v}{d t}\), so the equation becomes \(v+t \frac{d v}{d t}=f(v)\), which separates into \(\frac{d t}{t}=\frac{d v}{f(v)-v}\). Apply this method to solve the Euler-homogeneous \(DE\)s and \(IVP\)s in Problems \(41-44\). Plot sample solutions on a direction field and discuss. \(\frac{d y}{d t}=\frac{y^{2}+t y+t^{2}}{t^{2}}\) Equation Transcription: Text Transcription: y/t t/y dy/dt=f(y/t) v=y/t y=vt dy/dt=v+t dv/dt v+t dv/dt=f (v) d t/t=d v / f (v) -v DE IVP 41-44 dy/dt=y^2 + ty + t^2 / t^2

Another Conversion to Separable Equations Given the differential equation \(y^{\prime}=f(a t+b y+c)\), it can be shown that the substitution \(u=a t+b y+c\), where \(a\), \(b\), and \(c\) are constants, will transform the differential equation into the separable equation \(u^{\prime}=a+b f(u)\). Use this substitution to solve the \(DE\)s in Problems \(45-46\). \(y^{\prime}=(y+t)^{2}\) Equation Transcription: Text Transcription: y^prime=f (at+by+c) u=at+by+c a b c u^prime=a+bf (u) DE 45-46 y^prime=(y+t)^2

Another Conversion to Separable Equations Given the differential equation \(y^{\prime}=f(a t+b y+c)\), it can be shown that the substitution \(u=a t+b y+c\), where \(a\), \(b\), and \(c\) are constants, will transform the differential equation into the separable equation \(u^{\prime}=a+b f(u)\). Use this substitution to solve the \(DE\)s in Problems \(45-46\). \(y^{\prime}=e^{t+y-1}-1\) Equation Transcription: Text Transcription: y^prime=f (at+by+c) u=at+by+c a b c u^prime=a+bf (u) DE 45-46 y^prime=e^t+y-1 -1

Autonomous Equations Recall that if the right-hand side is independent of variable \(t\), so that the DE has the form \(y^{\prime}=f(y)\) the equation is called autonomous. (See Problem \(61\) in Sec. \(1.2\).) Such equations are always separable. (a) Identify the autonomous equations in Problems \(1-20\). (b) How can you recognize the direction field of an autonomous equation? Equation Transcription: Text Transcription: t y^prime=f(y) 61 1.2 1-20

Orthogonal Families Orthogonal Trajectories When a one-parameter family of curves satisfies a first-order \(DE\), we can find another such family as solution curves of a related \(DE\) with the property that a curve from one family intersects each curve of the other family orthogonally (that is, at right angles: their respective tangent lines are perpendicular). Each family constitutes the set of orthogonal trajectories for the other. For the following questions we use the customary independent variable \(x\) instead of \(t\). (a) Use implicit differentiation to show that the one-parameter family \(f(x, y)=c\) satisfies the differential equation \(d y / d x=-f_{x} / f_{y}\), where \(f_{x}=\partial f / \partial x\) and \(f_{y}=\partial f / \partial y\). (b) Explain why the curves satisfying \(d y / d x=f_{y} / f_{x}\) are the orthogonal trajectories to the family in part (a). (c) In Example 4, we found that the family \(x^{2}+y^{2}=c^{2}\) of circles with centers at the origin were the solution curves of the separable \(DE\) \(d y / d x=-x / y\). Use this and part (b) to show that the family of orthogonal trajectories are the straight lines \(y=k x\). (See Fig. \(1.3.7\). These families represent the electric field and equipotential lines around a point charge at the origin.) Equation Transcription: Text Transcription: DE x t f(x,y)=c dy/dx=-f_x / f_y f_x = partial f / partial x f_y = partial f / partial y dy/dx=f_y / f_x x^2+y^2=c^2 dy/dx=-x/y y=kx 1.3.7

More Orthogonal Trajectories Use Problem \(48\) to determine the family of trajectories orthogonal to each given family in Problems \(49-51\). (See Fig. \(1.3.8\)) \(y=c x^{2}\) Equation Transcription: Text Transcription: 48 49-51 1.3.8 y=cx^2

More Orthogonal Trajectories Use Problem \(48\) to determine the family of trajectories orthogonal to each given family in Problems \(49-51\). (See Fig. \(1.3.8\)) \(y=c / x^{2}\) Equation Transcription: Text Transcription: 48 49-51 1.3.8 y=c/x^2

More Orthogonal Trajectories Use Problem \(48\) to determine the family of trajectories orthogonal to each given family in Problems \(49-51\). (See Fig. \(1.3.8\)) \(x y=c\) Equation Transcription: Text Transcription: 48 49-51 1.3.8 xy=c

Calculator or Computer With the help of suitable computer software, for Problems \(52-55\) graph the families of curves along with their families of orthogonal trajectories. \(y=c\) (horizontal lines) Equation Transcription: Text Transcription: 52-55 y=c

Calculator or Computer With the help of suitable computer software, for Problems \(52-55\) graph the families of curves along with their families of orthogonal trajectories. \(4 x^{2}+y^{2}=c\) (ellipses) Equation Transcription: Text Transcription: 52-55 4x^2+y^2=c

Calculator or Computer With the help of suitable computer software, for Problems \(52-55\) graph the families of curves along with their families of orthogonal trajectories. \(x^{2}=4 c y^{3}\) (cubics) Equation Transcription: Text Transcription: 52-55 x^2=4cy^3

Calculator or Computer With the help of suitable computer software, for Problems \(52-55\) graph the families of curves along with their families of orthogonal trajectories. \(x^{2}+y^{2}=c y\) (coaxial circles) Equation Transcription: Text Transcription: 52-55 x^2+y^2=cy

The Sine Function The sine function \(sin x\) has the property that the square of itself plus the square of its derivative is identically equal to one. Find the most general function that has this property. Equation Transcription: Text Transcription: sin x

Disappearing Mothball The rate at which the volume of a mothball evaporates from solid to gas is proportional to the surface area of the ball. Suppose a mothball has been observed at one time to have a radius of \(0.5 in.\) and, six months later, a radius of \(0.25 in.\) (a) Express the radius of the mothball as a function of time. (b) When will the mothball disappear completely? Equation Transcription: Text Transcription: 0.5 in. 0.25 in.

Four-Bug Problem Four bugs sit at the comers of a square carpet \(L\) in. on a side. Simultaneously, each starts walking at the same rate of \(1 \text { in/sec }\) toward the bug on its right. See Fig. \(1.3.9\)(a). (a) Show that the bugs collide at the center of the carpet in exactly \(L\) sec. HINT: Each bug always moves in a direction perpendicular to the line of sight of the bug behind it, so the distance between two successive bugs always decreases at \(1 \text { in/sec }\). The bugs always form a square that is shrinking and rotating clockwise. (b) Using the result from (a), but using no calculus, tell how far each bug will travel. (c) Use differential equations to find the paths of the bugs. Simplify the setup by starting the bugs at the four points \((\pm 1,0)\) and \((0, \pm 1)\), making \(L \sqrt{2}\). Use Fig. \(1.3.9\)(b) to deduce the relationship \(d r \approx-r d \theta\), for sufficiently small \(d \theta\). Equation Transcription: Text Transcription: L 1 in/sec 1.3.9 (pm 1,0) (0,1 pm) L sqrt 2 dr approx -r d theta d theta

Radiant Energy Stefan’s Law of Radiation states that the radiation energy of a body is proportional to the fourth power of the absolute temperature \(T\) of a body. The rate of change of this energy in this surrounding medium of absolute temperature \(M\) is thus \(\frac{d T}{d t}=k\left(M^{4}-T^{4}\right)\), where \(k>0\) is a constant. Show that the general solution of Stefan’s equation is \(\ln \left|\frac{M+T}{M-T}\right|+2 \tan ^{-1}\left(\frac{T}{M}\right)=4 M^{3} k t+c\), where \(c\) is an arbitrary constant. Equation Transcription: Text Transcription: T M dT/dt=k(M^4-T^4) k>0 ln |M+T/M-T| +2 tan^-1 (T/M) = 4 M^3 kt+c c

Suggested Journal Entry Describe the distinction between quantitative and qualitative analysis. In what ways do they complement one another?