Question

1 Given that A=sinti+costj+ tk, evaluate ∣d^2A/dt^2∣
2 A particle moves along a curve whose parameter equations are x=e^−t, y=2cos3t, z=2sin3t. Find the magnitude of the acceleration at t=0
3 A particle moves along the curve x=2t^2, y=t^2−4t and z=3t−5,where t is the time. Find the components of the velocity at t=1 in the direction i−3j+2k
4 Determine the unit tangent at the point where t=2 on the curve x= t^2+1,y= 4t−3 and z=2t^2−6t
5 If A=5t^2+tj−t^3k and B=sinti−costj. evaluate d/dt (A⋅B)

Answer 1

Answer on Question #59205 – Math – Calculus

Question

1. Given that $A = \sin t i + \cos t j + t k$, evaluate $|d^2 A / dt^2|$

Solution

$$A = x i + y j + z k = \sin t \cdot i + \cos t \cdot j + t \cdot k,$$ $$\frac {d A}{d t} = \frac {d x}{d t} i + \frac {d y}{d t} j + \frac {d z}{d t} k,$$ $$\frac {d A}{d t} = \frac {d (\sin t)}{d t} i + \frac {d (\cos t)}{d t} j + \frac {d (t)}{d t} k = \cos t \cdot i - \sin t \cdot j + k,$$ $$\frac {d ^ {2} A}{d t ^ {2}} = \frac {d ^ {2} x}{d t ^ {2}} i + \frac {d ^ {2} y}{d t ^ {2}} j + \frac {d ^ {2} z}{d t ^ {2}} k = \frac {d}{d t} \left(\frac {d x}{d t}\right) i + \frac {d}{d t} \left(\frac {d y}{d t}\right) j + \frac {d}{d t} \left(\frac {d z}{d t}\right) k,$$ $$\frac {d ^ {2} A}{d t ^ {2}} = \frac {d}{d t} (\cos t) i + \frac {d}{d t} (- \sin t) j + \frac {d}{d t} (1) k = - \sin t \cdot i - \cos t \cdot j + 0 \cdot k = - \sin t \cdot i - \cos t \cdot j.$$

Answer: $\frac{d^2A}{dt^2} = -\sin t i - \cos t j.$

Question

2. A particle moves along a curve whose parameter equations are

$$x = e ^ {- t}, y = 2 \cos 3 t, z = 2 \sin 3 t.$$

Find the magnitude of the acceleration at $t = 0$.

Solution

The velocity of a particle:

$$v (t) = \frac {d x}{d t} i + \frac {d y}{d t} j + \frac {d z}{d t} k,$$ $$v (t) = \frac {d (e ^ {- t})}{d t} i + \frac {d (2 \cos 3 t)}{d t} j + \frac {d (2 \sin 3 t)}{d t} k = - e ^ {- t} i - 6 \sin 3 t \cdot j + 6 \cos 3 t \cdot k.$$

The acceleration of a particle:

$$a (t) = \frac {d ^ {2} x}{d t ^ {2}} i + \frac {d ^ {2} y}{d t ^ {2}} j + \frac {d ^ {2} z}{d t ^ {2}} k = \frac {d}{d t} \left(\frac {d x}{d t}\right) i + \frac {d}{d t} \left(\frac {d y}{d t}\right) j + \frac {d}{d t} \left(\frac {d z}{d t}\right) k,$$ $$a(t) = \frac{d}{dt}(-e^{-t})i + \frac{d}{dt}(-6\sin 3t)j + \frac{d}{dt}(6\cos 3t)k = e^{-t}i - 18\cos 3t \cdot j - 18\sin 3t \cdot k.$$

So, we have the acceleration at $t = 0$:

$$a(0) = e^{-0}i - 18\cos(3 \cdot 0) \cdot j - 18\sin(3 \cdot 0) \cdot k = i - 18j = a_xi + a_yj + a_zk.$$

The magnitude of the acceleration at $t = 0$:

$$|a(0)| = \sqrt{a_x^2 + a_y^2 + a_z^2},$$ $$|a(0)| = \sqrt{1 + 18^2} = \sqrt{325} = 5\sqrt{13}.$$

Answer: $|a(0)| = 5\sqrt{13}$.

Question

3. A particle moves along the curve

$$x = 2t^2, \quad y = t^2 - 4t, \quad z = 3t - 5,$$

where $t$ is time. Find the components of the velocity at $t = 1$ in the direction $i - 3j + 2k$

Solution

The velocity of a particle:

$$v(t) = v_xi + v_yj + v_zk = \frac{dx}{dt}i + \frac{dy}{dt}j + \frac{dz}{dt}k,$$ $$v(t) = \frac{d(2t^2)}{dt}i + \frac{d(t^2 - 4t)}{dt}j + \frac{d(3t - 5)}{dt}k = 4ti + (2t - 4)j + 3k = v_xi + v_yj + v_zk.$$

The vector projection of the velocity $v$ onto the direction $b = b_xi + b_yj + b_zk = i - 3j + 2k$:

$$pr_bv = \frac{v \cdot b}{b \cdot b} \cdot b = \frac{v_xb_x + v_yb_y + v_zb_z}{b_x^2 + b_y^2 + b_z^2} (b_xi + b_yj + b_zk),$$ $$pr_bv = \frac{4t \cdot 1 + (2t - 4) \cdot (-3) + 3 \cdot 2}{1^2 + (-3)^2 + 2^2} (i - 3j + 2k),$$ $$pr_bv = \frac{4t - 6t + 12 + 6}{14} (i - 3j + 2k),$$ $$pr_bv = \frac{-t + 9}{7} (i - 3j + 2k),$$ $$pr_bv(1) = \frac{-1 + 9}{7} (i - 3j + 2k),$$ $$pr_bv(1) = \frac{8}{7}i - \frac{24}{7}j + \frac{16}{7}k.$$

The components of the velocity at $t = 1$ in the given direction:

$$(pr_bv)_x = \frac{8}{7}; \quad (pr_bv)_y = -\frac{24}{7}; \quad (pr_bv)_z = \frac{16}{7}.$$

Answer: $(pr_b v)_x = \frac{8}{7}$; $(pr_b v)_y = -\frac{24}{7}$; $(pr_b v)_z = \frac{16}{7}$.

Question

4. Determine the unit tangent at the point where $t = 2$ on the curve

$$x = t^2 + 1, \quad y = 4t - 3, \quad z = 2t^2 - 6t$$

**Solution**

$$\begin{aligned} x'(t) &= \frac{dx}{dt} = \frac{d(t^2 + 1)}{dt} = 2t; \\ y'(t) &= \frac{dy}{dt} = \frac{d(4t - 3)}{dt} = 4; \\ z'(t) &= \frac{dz}{dt} = \frac{d(2t^2 - 6t)}{dt} = 4t - 6. \end{aligned}$$

The unit tangent:

$$\tau(t) = \frac{x'(t)}{\sqrt{x'^2(t) + y'^2(t) + z'^2(t)}} i + \frac{y'(t)}{\sqrt{x'^2(t) + y'^2(t) + z'^2(t)}} j + \frac{z'(t)}{\sqrt{x'^2(t) + y'^2(t) + z'^2(t)}} k,$$ $$\tau(t) = \frac{2t}{\sqrt{4t^2 + 16 + (4t - 6)^2}} i + \frac{4}{\sqrt{4t^2 + 16 + (4t - 6)^2}} j + \frac{4t - 6}{\sqrt{4t^2 + 16 + (4t - 6)^2}} k.$$

The unit tangent at the point where $t = 2$ on the curve:

$$\tau(2) = \frac{2 \cdot 2}{\sqrt{4 \cdot 2^2 + 16 + (4 \cdot 2 - 6)^2}} i + \frac{4}{\sqrt{4 \cdot 2^2 + 16 + (4 \cdot 2 - 6)^2}} j + \frac{4 \cdot 2 - 6}{\sqrt{4 \cdot 2^2 + 16 + (4 \cdot 2 - 6)^2}} k,$$ $$\tau(2) = \frac{4}{\sqrt{16 + 16 + 4}} i + \frac{4}{\sqrt{16 + 16 + 4}} j + \frac{4 \cdot 2 - 6}{\sqrt{16 + 16 + 4}} k,$$ $$\tau(2) = \frac{4}{6} i + \frac{4}{6} j + \frac{2}{6} k,$$ $$\tau(2) = \frac{2}{3} i + \frac{2}{3} j + \frac{1}{3} k,$$

Answer: $\tau(2) = \frac{2}{3} i + \frac{2}{3} j + \frac{1}{3} k$.

Question

5. If

$$A = 5t^2 i + tj - t^3 k \quad \text{and} \quad B = \sin t i - \cos t j$$

Evaluate $d/dt$ ($A \cdot B$).

Solution

If

$$A = A _ {x} i + A _ {y} j + A _ {z} k = 5 t ^ {2} i + t j - t ^ {3} k \quad \text{and}$$ $$B = B _ {x} i + B _ {y} j + B _ {z} k = \sin t \cdot i - \cos t \cdot j + 0 \cdot k$$

then

$$A \cdot B = A _ {x} B _ {x} + A _ {y} B _ {y} + A _ {z} B _ {z},$$ $$A \cdot B = 5 t ^ {2} \sin t + t \cdot (- \cos t) + (- t ^ {3} \cdot 0) = 5 t ^ {2} \sin t - t \cos t.$$ $$\begin{array}{l} \frac {d}{d t} (A \cdot B) = \frac {d}{d t} (5 t ^ {2} \sin t - t \cos t) = \\ = 10 t \sin t + 5 t ^ {2} \cos t - \cos t + t \sin t = 11 t \sin t + 5 t ^ {2} \cos t - \cos t. \\ \end{array}$$

Answer: $\frac{d}{dt} (A\cdot B) = 11t\sin t + 5t^2\cos t - \cos t.$

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