In each of Problems 1 through 4 sketch the trajectory corresponding to the solution satisfying the specified initial conditions, and indicate the direction of motion for increasing \( t \). 1. \( d x / d t=-x, \quad d y / d t=-2 y ; \quad x(0)=4, \quad y(0)=2 \) 2. \( d x / d t=-x, \quad d y / d t=2 y ; \quad x(0)=4, \quad y(0)=2 \quad \) and \( \quad x(0)=4, \quad y(0)=0 \) 3. \( d x / d t=-y, \quad d y / d t=x ; \quad x(0)=4, \quad y(0)=0 \quad \) and \( \quad x(0)=0, \quad y(0)=4 \) 4. \( d x / d t=a y, \quad d y / d t=-b x, \quad a>0, \quad b>0 ; \quad x(0)=\sqrt{a}, \quad y(0)=0 \) For each of the systems in Problems 5 through 16:
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Step 1: Given the system of differential equations for problem 3: \[ \frac{dx}{dt} = -y, \quad \frac{dy}{dt} = x \] with initial conditions \( x(0) = 4, y(0) = 0 \) and \( x(0) = 0, y(0) = 4 \). Show more…
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Accordiing to the problem So, $$ \begin{gathered} \vec{w}=w(-\vec{j}) \\ w_{x}=\frac{d v_{x}}{d t}=0 \text { and } w_{y}=\frac{d v_{y}}{d t}=-w \end{gathered} $$ Differentiating Eq. of trajectory, $y=a x-b x^{2}$, with respect to time $$ \frac{d y}{d t}=\frac{a d x}{d t}-2 b x \frac{d x}{d t} $$ So, $$ \left.\frac{d y}{d t}\right|_{x=0}=\left.a \frac{d x}{d t}\right|_{x=0} $$ Again differentiating with respect to time or, $$ \begin{gathered} \frac{d^{2} y}{d t^{2}}=\frac{a d^{2} x}{d t^{2}}-2 b\left(\frac{d x}{d t}\right)^{2}-2 b x \frac{d^{2} x}{d t^{2}} \\ -w=a(0)-2 b\left(\frac{d x}{d t}\right)^{2}-2 b x(0)(\text { using } 1) \end{gathered} $$ or, $$ \frac{d x}{d t}=\sqrt{\frac{w}{2 b}} \text { (using 1) } $$ Using (3) in (2) $\left.\quad \frac{d y}{d t}\right|_{x=0}=a \sqrt{\frac{w}{2 b}}$ Hence, the velocity of the particle at the origin $v=\sqrt{\left(\frac{d x}{d t}\right)_{x=0}^{2}+\left(\frac{d y}{d t}\right)_{x=0}^{2}}=\sqrt{\frac{w}{2 b}+a^{2} \frac{w}{2 b}}$ (using Eqns (3) and (4)) Hence, $v=\sqrt{\frac{w}{2 b}\left(1+a^{2}\right)}$
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The $x$ - and $y$ -coordinates of a projectile launched from the origin are $x=v_{0 x} t$ and $y=v_{0 y} t-\frac{1}{2} g t^{2}$ . Solve the first of these equations for time $t$ and substitute into the second to show that the path of a projectile is a parabola with the form $y=a x+b x^{2},$ where $a$ and $b$ are constants.
Where trajectories crest $\quad$ For a projectile fired from the ground at launch angle $\alpha$ with initial speed $v_{0},$ consider $\alpha$ as a variable and $v_{0}$ as a fixed constant. For each $\alpha, 0<\alpha<\pi / 2,$ we obtain a parabolic trajectory as shown in the accompanying figure. Show that the points in the plane that give the maximum heights of these parabolic trajectories all lie on the ellipse $$x^{2}+4\left(y-\frac{v_{0}^{2}}{4 g}\right)^{2}=\frac{v_{0}^{4}}{4 g^{2}}$$ where $x \geq 0.$ (GRAPH CANNOT COPY)
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