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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 Frank Black. in Pakistan
Jan 10,2025

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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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To solve the given problem regarding the vector functions \( \vec{F}(t) \) and \( \vec{G}(t) \), it's essential to have a solid understanding of vector calculus. The operations involved, like the cross product and differentiation, are foundational techniques in fields such as physics and engineering, where they're used to describe rotational dynamics and spatial interactions between objects in motion. Furthermore, the properties of the cross product are particularly fascinating! If two vectors are perpendicular, their cross product yields a vector that is at a right angle to both. This means that understanding the geometric interpretation of vector operations not only strengthens mathematical insight but also enhances problem-solving skills in real-world scenarios, such as when analyzing forces acting on an object at rest or in motion.

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