answer:The position vector is given as: {eq}r(t)=left langle sin 7t,cos 6t,cos 3t right rangle {/eq} Thus, the velocity of the particle is: {eq}v(t)=r'(t)=left langle 7cos 7t, -6sin 6t, -3sin 3t right rangle {/eq} Therefore, the velocity at {eq}t=frac{pi }{2} {/eq} is: {eq}begin{align*} vleft ( frac{pi}{2} right ) & = left langle 7cos left ( frac{7pi}{2} right ), -6sin left ( frac{6pi}{2} right ), -3sin left ( frac{3pi}{2} right ) right rangle & = left langle 7(0), -6(0), -3(-1) right rangle & = left langle 0, 0, 3 right rangle end{align*} {/eq} Hence, the velocity of the particle at {eq}t=frac{pi}{2} {/eq} is {eq}left langle 0, 0, 3 right rangle {/eq}.

answer:To multiply these two vectors, you perform the dot product for each row of the first vector with the column of the second vector. The result is a new vector with the products. Here's the calculation: First row: -frac{5}{2} cdot (-3) + 2 cdot 1 = frac{15}{2} - frac{2}{2} = frac{15}{2} - frac{2}{2} = frac{13}{2} Second row: 1 cdot (-3) + 1 cdot 1 = -3 + 1 = -2 Since the second vector is a 1x2 matrix, the result will be a 3x1 matrix with the calculated values: left( begin{array}{c} frac{13}{2} -2 1 cdot (-3) + 1 cdot 1 end{array} right) = left( begin{array}{c} frac{13}{2} -2 -2 end{array} right) So the product of these two vectors is: left( begin{array}{c} frac{13}{2} -2 -2 end{array} right)

answer:We have that: |y'''(x)| = |(xy''(x) + y'(x))cos(xy) - (xy'(x) + y(x))sin(xy)(y'(x) + xy''(x))| From the previous analysis, we know that |y'(x)| leq 3 and |y''(x)| leq 3 + |y'(x)| leq 6. Using a numerical method, we can approximate y(3) and find an upper bound for |y(x)|, say |y(x)| leq M. Then, we have: |y'''(x)| leq (6M + 3)|cos(xy)| + (3M + 6)(3 + 6)|sin(xy)| leq 9(6M + 3) + 9(3M + 6) = 81M + 81 Therefore, a bound for |y'''(x)| is 81M + 81, where M is an upper bound for |y(x)|.

answer:The solution to the equation is (x = -4.1). Here's a step-by-step explanation: begin{align*} -2.8 &= x + 1.3 x &= -2.8 - 1.3 x &= -4.1 end{align*}