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Question If \( \vec{F}(t)=\underline{i}+2 e^{2 t} \underline{j}+t^{3} \underline{k} \) and \( \vec{G}(t)=3 t^{2} \underline{i}+5 e^{-t} \underline{j}-t 3 \underline{k} \) are the vector functions then evaluate OR a) \( \frac{d}{d t}(\vec{F} \times \vec{G})(t) \) b) \( \frac{d \vec{F}}{d t} \times \vec{G} \) c) \( \vec{F} \times \frac{d \vec{G}}{d t} \) and verify d) \( \frac{d}{d t}(\vec{F} \times \vec{G})(t)=\frac{d \vec{F}}{d t} \times \vec{G}+\vec{F} \times \frac{d \vec{G}}{d t} \) (a) If \( A(-2,5), B(1,5) \) are end points of chord \( A B \) of circle \( x^{2}+y^{2}+x-5 y-2=0 \), then show that line drawn from the center of circle is perpendicular to chord \( A B \), and bisects chord \( A B \). (b) Coordinates of end points of two chords are \( P(0,2), Q(-2,0) \) and \( R(0,-2), S(2,0) \). Show that the two chords \( P Q \) and \( R S \) are equidistant from the center of circle \( x^{2}+y^{2}=4 \). Solve the differential equation \[ y^{2} d x+\left(x y+x^{2}\right) d y=0 \]

Ask by Welch Ryan. in Pakistan
Jan 10,2025

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Tutor-Verified Answer

Answer

The solution to the differential equation \( y^{2} d x+\left(x y+x^{2}\right) d y=0 \) is: \[ \frac{y^{3}}{3} + \frac{x^{2}y}{2} + \frac{x^{3}}{3} = C \] where \( C \) is a constant.

Solution

To solve the given differential equation \( y^{2} d x+\left(x y+x^{2}\right) d y=0 \), we can use the method of separation of variables. First, let's rearrange the equation to separate the variables: \[ y^{2} d x = -(x y+x^{2}) d y \] Next, we can integrate both sides of the equation: \[ \int y^{2} d x = -\int (x y+x^{2}) d y \] Now, we can solve the integrals: \[ \frac{y^{3}}{3} = -\frac{x^{2}y}{2} - \frac{x^{3}}{3} + C \] where \( C \) is the constant of integration. Therefore, the solution to the differential equation is: \[ \frac{y^{3}}{3} + \frac{x^{2}y}{2} + \frac{x^{3}}{3} = C \] This is the general solution to the differential equation.

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Beyond the Answer

The fascinating world of vectors and their properties can be traced back to the early days of physics and mathematics. In the 19th century, mathematicians like William Rowan Hamilton began developing vector analysis, laying the groundwork for concepts used today. Vectors make understanding complex dynamics in physics and engineering simpler, helping us think in multidimensional spaces—a game-changer for calculus! Now, onto practical applications! The cross product of vectors, like \( \vec{F} \) and \( \vec{G} \), represents torque and angular momentum in physics. This means when you're dealing with anything that spins or rotates, the calculations often involve cross products. Whether you're building a robot arm or analyzing planetary motion, grasping these vector functions boosts your problem-solving game tremendously!

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