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 \[ y^{2} d x+\left(x y+x^{2}\right) d y=0 \]
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To tackle this vector calculus problem, it’s helpful to remember a fundamental rule from vector theory: the derivative of a cross product follows the product rule! This means that the derivative of the cross product of two vectors in time is the derivative of the first vector crossed with the second plus the first vector crossed with the derivative of the second. So for \( \frac{d}{d t}(\vec{F} \times \vec{G}) \), use this rule! Now, as for verifying that the line from the center of the circle is perpendicular to chord \( AB \), we can start by finding the center and radius from the given equation. Once we obtain the center, we can calculate the slope of the chord \( AB \) formed by the endpoints \( A(-2,5) \) and \( B(1,5) \), and the slope of the radius from the center to the midpoint of chord \( AB \). If these slopes multiply to -1, then you've confirmed that it’s indeed perpendicular!
