(6 marks) QUESTION TWO (15 Marks) (a) A particle moves along the curve x=t, y=t^2 and z=t^3 where t is the time. Find the components of its velocity and acceleration at time t=1 (b) The acceleration of a particle is given by (5 marks) (d^2overrightarrow (r))/(dt^2)=(3cos(3t))hat (i)+(3sin(3t))hat (j)+that (k) If overrightarrow (V) is zero at t=0 , find overrightarrow (V) at any time. (5 marks) (c) Find the angle between A=2i+2j-k and

Let's solve each part of the question step by step.<br /><br />### Part (a)<br />A particle moves along the curve \( x = t \), \( y = t^2 \), and \( z = t^3 \) where \( t \) is the time. We need to find the components of its velocity and acceleration at time \( t = 1 \).<br /><br />**Velocity:**<br /><br />The velocity vector \(\overrightarrow{v}\) is given by the derivative of the position vector \(\overrightarrow{r}(t)\) with respect to time \( t \).<br /><br />\[<br />\overrightarrow{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k} = t\hat{i} + t^2\hat{j} + t^3\hat{k}<br />\]<br /><br />Taking the derivative with respect to \( t \):<br /><br />\[<br />\overrightarrow{v}(t) = \frac{d\overrightarrow{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} + \frac{dz}{dt}\hat{k}<br />\]<br /><br />\[<br />\overrightarrow{v}(t) = \frac{d(t)}{dt}\hat{i} + \frac{d(t^2)}{dt}\hat{j} + \frac{d(t^3)}{dt}\hat{k}<br />\]<br /><br />\[<br />\overrightarrow{v}(t) = 1\hat{i} + 2t\hat{j} + 3t^2\hat{k}<br />\]<br /><br />At \( t = 1 \):<br /><br />\[<br />\overrightarrow{v}(1) = 1\hat{i} + 2(1)\hat{j} + 3(1)^2\hat{k} = 1\hat{i} + 2\hat{j} + 3\hat{k}<br />\]<br /><br />So, the components of the velocity at \( t = 1 \) are \( 1 \), \( 2 \), and \( 3 \).<br /><br />**Acceleration:**<br /><br />The acceleration vector \(\overrightarrow{a}\) is given by the derivative of the velocity vector \(\overrightarrow{v}(t)\) with respect to time \( t \).<br /><br />\[<br />\overrightarrow{a}(t) = \frac{d\overrightarrow{v}}{dt} = \frac{d(1)}{dt}\hat{i} + \frac{d(2t)}{dt}\hat{j} + \frac{d(3t^2)}{dt}\hat{k}<br />\]<br /><br />\[<br />\overrightarrow{a}(t) = 0\hat{i} + 2\hat{j} + 6t\hat{k}<br />\]<br /><br />At \( t = 1 \):<br /><br />\[<br />\overrightarrow{a}(1) = 0\hat{i} + 2\hat{j} + 6(1)\hat{k} = 0\hat{i} + 2\hat{j} + 6\hat{k}<br />\]<br /><br />So, the components of the acceleration at \( t = 1 \) are \( 0 \), \( 2 \), and \( 6 \).<br /><br />### Part (b)<br />The acceleration of a particle is given by:<br /><br />\[<br />\frac{d^2\overrightarrow{r}}{dt^2} = (3\cos(3t))\hat{i} + (3\sin(3t))\hat{j} + t\hat{k}<br />\]<br /><br />If \(\overrightarrow{V}\) is zero at \( t = 0 \), we need to find \(\overrightarrow{V}\) at any time \( t \).<br /><br />To find \(\overrightarrow{V}(t)\), we integrate the acceleration vector with respect to time \( t \):<br /><br />\[<br />\overrightarrow{V}(t) = \int \left[ (3\cos(3t))\hat{i} + (3\sin(3t))\hat{j} + t\hat{k} \right] dt<br />\]<br /><br />Integrating each component separately:<br /><br />\[<br />\overrightarrow{V}(t) = \int 3\cos(3t) dt \hat{i} + \int 3\sin(3t) dt \hat{j} + \int t dt \hat{k}<br />\]<br /><br />\[<br />\overrightarrow{V}(t) = \left( \frac{3}{3}\sin(3t) \right) \hat{i} + \left( -\frac{3}{3}\cos(3t) \right) \hat{j} + \left( \frac{t^2}{2} \right) \hat{k} + \overrightarrow{C}<br />\]<br /><br />\[<br />\overrightarrow{V}(t) = \sin(3t)\hat{i} - \cos(3t)\hat{j} + \frac{t^2}{2}\hat{k} + \overrightarrow{C}<br />\]<br /><br />Given that \(\overrightarrow{V}(0) = 0\):<br /><br />\[<br />\overrightarrow{V}(0) = \sin(0)\hat{i} - \cos(0)\hat{j} + \frac{0^2}{2}\hat{k} + \overrightarrow{C} = 0\hat{i} - 1\hat{j} + 0\hat{k} + \overrightarrow{C} = 0<br />\]<br /><br />\[<br />\overrightarrow{C} = \hat{j}<br />\]<br /><br />Thus,<br /><br />\[<br />\overrightarrow{V}(t) = \sin(3t)\hat{i} - \cos(3t)\hat{j} + \frac{t^2}{2}\hat{k} + \hat{j}<br />\]<br /><br />\[<br />\overrightarrow{V}(t) = \sin(3t)\hat{i} - (\cos(3t) - 1)\hat{j} + \frac{t^2}{2}\hat{k}<br />\]<br /><br />### Part (c)<br />Find the angle between \(\overrightarrow{A} = 2\hat{i} + 2\hat{j} - \hat{k}\) and \(\overrightarrow{B}\).<br /><br />First, we need the dot product of \(\overrightarrow{A}\) and \(\overrightarrow{B}\):<br /><br />\[<br />\overrightarrow{A} \cdot \overrightarrow{B} = (2)(B_x) + (2)(B_y) + (-1)(B_z)<br />\]<br /><br />Next, we need the magnitudes of \(\overrightarrow{A}\) and \(\overrightarrow{B}\):<br /><br />\[<br />|\overrightarrow{A}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3<br />\]<br /><br />\[<br />|\overrightarrow{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2}<br />\]<br /><br />The cosine of the angle \(\theta\) between \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is given by:<br /><br />\[<br />\cos(\theta) = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}<br />\]<br /><br />Since \(\overrightarrow{B}\) is not specified, we cannot compute the exact angle without knowing \(\overrightarrow{B}\). If \(\overrightarrow{B}\) were provided, we would substitute the values into the formula above to find \(\theta\).<br /><br />Please provide the components of \(\overrightarrow{B}\) if you want the exact angle calculation.