Question 1

The domain of the vector function $\mathbf { v } ( t ) = e ^ { t } \mathbf { i } - \sqrt { 1 - t ^ { 2 } } \mathbf { j } + 8 \mathbf { k }$ is
A) $[ - 1,1 )$
B) $( - 1,1 )$
C) $[ - 1,1 ]$
D) $( - 1,1 ]$
E) $( - \infty , \infty )$

Accepted Answer

The domain is determined by the square root function $\sqrt{1 - t^2}$, which requires $1 - t^2 \geq 0$. Solving this inequality gives $-1 \leq t \leq 1$, so the domain is $[-1, 1]$. The exponential function $e^t$ and the constant function $8$ do not restrict the domain further.

Question 2

The domain of the vector function $\mathbf { v } ( t ) = 5 t ^ { 2 } \mathbf { i } - \ln ( 3 t - 4 ) \mathbf { j } + \sin t \mathbf { k }$ is
A) $\left( - \frac { 4 } { 3 } , \infty \right)$
B) $\left( \frac { 4 } { 3 } , \infty \right)$
C) $\left( - \frac { 4 } { 3 } , \frac { 4 } { 3 } \right)$
D) $\left( - \infty , \frac { 4 } { 3 } \right)$
E) $( - \infty , \infty )$

Accepted Answer

The domain of the vector function is determined by the restrictions imposed by its components. The first component, $5t^2$, is defined for all real numbers. The third component, $\sin t$, is also defined for all real numbers. The second component, $-\ln(3t - 4)$, requires that $3t - 4 > 0$, which simplifies to $t > \frac{4}{3}$. Therefore, the domain of the entire vector function is $\left( \frac{4}{3}, \infty \right)$.

Question 3

The domain of the vector function $\mathbf { v } ( t ) = \frac { 3 } { \sqrt { t ^ { 2 } - 4 } } \mathbf { i } - \sin ( 2 t ) \mathbf { j } + \frac { 1 } { t } \mathbf { k }$ is
A) $( - \infty , - 2 )$
B) $( 2 , \infty )$
C) $( - 2,2 )$
D) $( - \infty , - 2 ) \cup ( 2 , \infty )$
E) $( - \infty , \infty )$

Accepted Answer

The domain of the vector function is determined by the values of $t$ for which the function is defined. The denominator in the first and third components of the vector function cannot be zero. For the first component, $t^2 - 4 > 0$, which is true for $t < -2$ and $t > 2$. For the third component, $t \neq 0$. Since $\sin(2t)$ is defined for all real numbers, it does not restrict the domain further. Therefore, the domain is $( - \infty , - 2 ) \cup ( 2 , \infty )$.

Question 4

The domain of the vector function $\mathbf { v } ( t ) = 2 ^ { t } \mathbf { i } - 5 t ^ { 2 } \mathbf { j } + \log _ { 2 } ( t - 4 ) \mathbf { k }$ is
A) $( 4 , \infty )$
B) $( - 4 , \infty )$
C) $( - \infty , - 4 )$
D) $( - \infty , - 4 ) \cup ( 4 , \infty )$
E) $( - 4,4 )$

Accepted Answer

The domain is determined by the restrictions on $t$ from each component of the vector function. $2^t$ is defined for all real numbers, and $-5t^2$ is also defined for all real numbers. The restriction comes from $\log_2(t-4)$, which requires $t-4 > 0$, or $t > 4$. Therefore, the domain is $(4, \infty)$.

Question 5

Let $\mathbf { r } ( t ) = \cos ( 2 t - 1 ) \mathbf { i } - \sqrt { t ^ { 2 } - 1 } \mathbf { j } + \frac { 1 } { 1 + t } \mathbf { k } \text {. Then }$ is
A) $2 \sin ( 2 t - 1 ) \mathbf { i } + \frac { t } { \sqrt { t ^ { 2 } - 1 } } \mathbf { j } - \frac { 1 } { ( 1 + t ) ^ { 2 } } \mathbf { k }$
B) $- 2 \sin ( 2 t - 1 ) \mathbf { i } + \frac { t } { \sqrt { t ^ { 2 } - 1 } } \mathbf { j } + \frac { 1 } { ( 1 + t ) ^ { 2 } } \mathbf { k }$
C) $- 2 \sin ( 2 t - 1 ) \mathbf { i } + \frac { t } { \sqrt { t ^ { 2 } - 1 } } \mathbf { j } - \frac { 1 } { ( 1 + t ) ^ { 2 } } \mathbf { k }$
D) $- 2 \sin ( 2 t - 1 ) \mathbf { i } - \frac { t } { \sqrt { t ^ { 2 } - 1 } } \mathbf { j } - \frac { 1 } { ( 1 + t ) ^ { 2 } } \mathbf { k }$
E) $2 \sin ( 2 t - 1 ) \mathbf { i } - \frac { t } { \sqrt { t ^ { 2 } - 1 } } \mathbf { j } - \frac { 1 } { ( 1 + t ) ^ { 2 } } \mathbf { k }$

Accepted Answer

The answer of Let \(\mathbf { r } ( t...

Question 6

Let $\mathbf { r } ( t ) = e ^ { \cos t } \mathbf { i } + \tan t \mathbf { j } + \cos ( 5 - t ) \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( t )$ is
A) $\sin t e ^ { \cos t } \mathbf { i } + \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
B) $- \sin t e ^ { \cos t } \mathbf { i } + \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
C) $- \sin t e ^ { \cos t } \mathbf { i } - \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
D) $\sin t e ^ { \cos t } \mathbf { i } - \sec ^ { 2 } t \mathbf { j } + \sin ( 5 - t ) \mathbf { k }$
E) $\sin t e ^ { \cos t } \mathbf { i } + \sec ^ { 2 } t \mathbf { j } - \sin ( 5 - t ) \mathbf { k }$

Accepted Answer

To find $\mathbf { r } ^ { \prime } ( t )$, we differentiate each component of $\mathbf { r } ( t )$ with respect to $t$:- For the $\mathbf{i}$ component, the derivative of $e^{\cos t}$ is $-\sin t e^{\cos t}$ using the chain rule.- For the $\mathbf{j}$ component, the derivative of $\tan t$ is $\sec^2 t$.- For the $\mathbf{k}$ component, the derivative of $\cos(5 - t)$ is $\sin(5 - t)$ but with a negative sign due to the chain rule, making it $\sin(t - 5)$, which is equivalent to $\sin(5 - t)$ in magnitude but opposite in sign. However, the correct sign is positive as in the original statement, indicating a mistake in my explanation regarding the sign. The correct derivative, considering the chain rule correctly, should indeed result in a positive sign for the $\mathbf{k}$ component, aligning with the options provided. Thus, the correct derivative is $\sin(5 - t)$, maintaining the positive sign as per the chain rule applied correctly to the derivative of $\cos(5 - t)$, which involves multiplying by the derivative of the inside function, $-1$, resulting in $-\sin(5 - t)$, but this contradicts the correct interpretation of the derivative as presented in the options. The correct interpretation should acknowledge the derivative of $\cos(5 - t)$ as $\sin(5 - t)$ with a negative sign due to the derivative of $-t$ being $-1$, which correctly results in $-\sin(5 - t)$, aligning with the principles of differentiation and the chain rule.

Question 7

Let $\mathbf { r } ( t ) = - \sin t \mathbf { i } + \cos t \mathbf { j } - e ^ { - t } \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( t )$ is
A) $- \cos t \mathbf { i } + \sin t \mathbf { j } + e ^ { - t } \mathbf { k }$
B) $- \cos t \mathbf { i } - \sin t \mathbf { j } - e ^ { - t } \mathbf { k }$
C) $\cos t \mathbf { i } - \sin t \mathbf { j } + e ^ { - t } \mathbf { k }$
D) $- \cos t \mathbf { i } - \sin t \mathbf { j } + e ^ { - t } \mathbf { k }$
E) $\cos t \mathbf { i } - \sin t \mathbf { j } - e ^ { - t } \mathbf { k }$

Accepted Answer

To find $\mathbf { r } ^ { \prime } ( t )$, we differentiate each component of $\mathbf { r } ( t )$ with respect to $t$. The derivative of $-\sin t$ with respect to $t$ is $-\cos t$, the derivative of $\cos t$ with respect to $t$ is $-\sin t$, and the derivative of $-e^{-t}$ with respect to $t$ is $e^{-t}$. Therefore, $\mathbf { r } ^ { \prime } ( t ) = - \cos t \mathbf { i } - \sin t \mathbf { j } + e ^ { - t } \mathbf { k }$.

Question 8

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + e ^ { \cos t } \mathbf { j } - \sec t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( t )$ is
A) $\cos t e ^ { \sin t } \mathbf { i } - \sin t e ^ { \cos t } \mathbf { j } - \sec t \tan t \mathbf { k }$
B) $\cos t e ^ { \sin t } \mathbf { i } - \sin t e ^ { \cos t } \mathbf { j } + \sec t \tan t \mathbf { k }$
C) $\cos t e ^ { \sin t } \mathbf { i } + \sin t e ^ { \operatorname { cost } } \mathbf { j } - \sec t \tan t \mathbf { k }$
D) $\cos t e ^ { \sin t } \mathbf { i } + \sin t e ^ { \operatorname { cost } t } \mathbf { j } + \sec t \tan t \mathbf { k }$
E) $- \cos t e ^ { \sin t } \mathbf { i } - \sin t e ^ { \cos t } \mathbf { j } - \sec t \tan t \mathbf { k }$

Accepted Answer

The derivative of $\mathbf { r } ( t )$ is$$\mathbf { r } ^ { \prime } ( t ) = e ^ { \sin t } \cos t \mathbf { i } - e ^ { \cos t } \sin t \mathbf { j } - \sec t \tan t \mathbf { k }$$

Question 9

Let $\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is
A) $4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$
B) $4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$
C) $- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$
D) $- 4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$
E) $- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } - \mathbf { k }$

Accepted Answer

The answer of Let \(\mathbf { r } ( t...

Question 10

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + \cos t \mathbf { j } + \ln ( 5 - t ) \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

A) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + \cos t \mathbf { j } - \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }$
B) $e ^ { \operatorname { tin } t } \left( \cos ^ { 2 } - \sin t \right) \mathbf { i } - \cos t \mathbf { j } + \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }$
C) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + \cos t \mathbf { j } + \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }$
D) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } - \cos t \mathbf { j } - \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }$
E) $- e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } - \cos t \mathbf { j } + \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }$

Accepted Answer

The answer of Let \(\mathbf { r } ( t...

Question 11

Let $\mathbf { r } ( t ) = - \cos t \mathbf { i } + \sin t \mathbf { j } - e ^ { - t } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

A) $- \cos t \mathbf { i } - \sin t \mathbf { j } + e ^ { - t } \mathbf { k }$
B) $\cos t \mathbf { i } - \sin t \mathbf { j } + e ^ { - t } \mathbf { k }$
C) $\cos t \mathbf { i } + \sin t \mathbf { j } + e ^ { - t } \mathbf { k }$
D) $- \cos t \mathbf { i } - \sin t \mathbf { j } - e ^ { - t } \mathbf { k }$
E) $\cos t \mathbf { i } - \sin t \mathbf { j } - e ^ { - t } \mathbf { k }$

Accepted Answer

The answer of Let \(\mathbf { r } ( t...

Question 12

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } - e ^ { \cos t } \mathbf { j } + \ln t \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

A) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
B) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
C) $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
D) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } - e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$
E) $e ^ { \sin t } \left( \cos ^ { 2 } t + \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t + \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$

Accepted Answer

To find $\mathbf { r } ^ { \prime \prime } ( t )$, we first differentiate $\mathbf { r } ( t )$ to get $\mathbf { r } ^ { \prime } ( t )$, and then differentiate $\mathbf { r } ^ { \prime } ( t )$ again. For the $\mathbf{i}$ component, differentiating $e^{\sin t}$ once gives $e^{\sin t} \cos t$, and differentiating again gives $e^{\sin t}(\cos^2 t - \sin t)$ due to the product rule and chain rule.For the $\mathbf{j}$ component, differentiating $-e^{\cos t}$ once gives $e^{\cos t} \sin t$, and differentiating again gives $e^{\cos t}(\cos t - \sin^2 t)$, applying the product rule and chain rule similarly.For the $\mathbf{k}$ component, differentiating $\ln t$ once gives $\frac{1}{t}$, and differentiating again gives $-\frac{1}{t^2}$.Thus, the correct answer is $e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + e ^ { \cos t } \left( \cos t - \sin ^ { 2 } t \right) \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$, which matches choice A.

Question 13

Let $\mathbf { v } ( t ) = e ^ { 3 t } \mathbf { i } + e ^ { - 3 t } \mathbf { j } + t \mathbf { k }$ and $\mathbf { w } ( t ) = e ^ { - 2 t } \mathbf { i } + e ^ { 2 t } \mathbf { j } - t \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \bullet \mathbf { w } ( t ) )$ is

A) $e ^ { t } - e ^ { - t } + 2 t$
B) $e ^ { t } + e ^ { - t } - 2 t$
C) $e ^ { t } - e ^ { - t } - 2 t$
D) $e ^ { t } + e ^ { - t } + 2 t$
E) $- e ^ { t } - e ^ { - t } + 2 t$

Accepted Answer

The answer of Let \(\mathbf { v } ( t...

Question 14

Let $\mathbf { v } ( t ) = \sin t \mathbf { i } + \cos t \mathbf { j } + t \mathbf { k }$ and $\mathbf { w } ( t ) = \sin t \mathbf { i } + \cos t \mathbf { j } + t ^ { 3 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \bullet \mathbf { w } ( t ) )$ is

A) $4 t ^ { 3 }$
B) $4 t ^ { 5 }$
C) $- 4 t ^ { 3 }$
D) $- 3 t ^ { 3 }$
E) $3 t ^ { 3 }$

Accepted Answer

The dot product $\mathbf { v } ( t ) \bullet \mathbf { w } ( t )$ is computed first: $(\sin t)(\sin t) + (\cos t)(\cos t) + (t)(t^3) = 1 + t^4$, since $\sin^2 t + \cos^2 t = 1$. Differentiating $1 + t^4$ with respect to $t$ gives $4t^3$.

Question 15

Let $\mathbf { v } ( t ) = \sin t \mathbf { i } - \cos t \mathbf { j } + 3 t ^ { 2 } \mathbf { k }$ and $\mathbf { w } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t ^ { 3 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \cdot \mathbf { w } ( t ) )$ is

A) $4 t ^ { 3 }$
B) $- 5 t ^ { 4 }$
C) $15 t ^ { 4 }$
D) $5 t ^ { 4 }$
E) $- 15 t ^ { 4 }$

Accepted Answer

The answer of Let \(\mathbf { v } ( t...

Question 16

Let $\mathbf { v } ( t ) = \sin t \mathbf { i } - 2 \cos t \mathbf { j } + t ^ { 2 } \mathbf { k }$ and $\mathbf { w } ( t ) = 3 \cos t \mathbf { i } + \sin t \mathbf { j } - t ^ { 2 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \bullet \mathbf { w } ( t ) )$ is

A) $\cos ( 2 t ) + 4 t ^ { 3 }$
B) $\cos ( 2 t ) - 4 t ^ { 3 }$
C) $- \cos ( 2 t ) - 4 t ^ { 3 }$
D) $- \cos ( 2 t ) + 4 t ^ { 3 }$
E) $2 \cos ( 2 t ) - 4 t ^ { 3 }$

Accepted Answer

To find $\frac { d } { d t } ( \mathbf { v } ( t ) \bullet \mathbf { w } ( t ) )$, first compute the dot product $\mathbf { v } ( t ) \bullet \mathbf { w } ( t )$, then differentiate the result with respect to $t$. The dot product is given by:$$\mathbf { v } ( t ) \bullet \mathbf { w } ( t ) = (\sin t)(3\cos t) + (-2\cos t)(\sin t) + (t^2)(-t^2) = 3\sin t \cos t - 2\sin t \cos t - t^4 = \sin t \cos t - t^4.$$Using the trigonometric identity $\sin(2t) = 2\sin t \cos t$, we can rewrite $\sin t \cos t$ as $\frac{1}{2}\sin(2t)$. Thus, the dot product simplifies to $\frac{1}{2}\sin(2t) - t^4$.Differentiating this with respect to $t$ gives:$$\frac{d}{dt}\left(\frac{1}{2}\sin(2t) - t^4\right) = \cos(2t) - 4t^3.$$Therefore, the correct answer is $\cos(2t) - 4t^3$, which corresponds to choice B.

Question 17

Let $\mathbf { v } ( t ) = t ^ { 2 } \mathbf { i } - \mathbf { j } + 3 t \mathbf { k }$ and $\mathbf { w } ( t ) = 2 t \mathbf { i } + 3 \mathbf { j } - t \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

A) $- 8 \mathbf { i } - \left( 3 t ^ { 2 } + 12 t \right) \mathbf { j } + ( 6 t + 2 ) \mathbf { k }$
B) $8 \mathbf { i } + \left( 3 t ^ { 2 } + 12 t \right) \mathbf { j } - ( 6 t + 2 ) \mathbf { k }$
C) $8 \mathbf { i } + \left( 3 t ^ { 2 } + 12 t \right) \mathbf { j } + ( 6 t + 2 ) \mathbf { k }$
D) $- 8 \mathbf { i } + \left( 3 t ^ { 2 } + 12 t \right) \mathbf { j } + ( 6 t + 2 ) \mathbf { k }$
E) $- 8 \mathbf { i } + \left( 3 t ^ { 2 } + 12 t \right) \mathbf { j } - ( 6 t + 2 ) \mathbf { k }$

Accepted Answer

The derivative of the cross product of two vectors can be found using the product rule for derivatives, which states $\frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \frac{d\mathbf{u}}{dt} \times \mathbf{v} + \mathbf{u} \times \frac{d\mathbf{v}}{dt}$. Applying this rule to $\mathbf{v}(t) \times \mathbf{w}(t)$ and calculating the cross products and their derivatives yields the vector $- 8 \mathbf { i } + \left( 3 t ^ { 2 } + 12 t \right) \mathbf { j } + ( 6 t + 2 ) \mathbf { k }$.

Question 18

Let $\mathbf { v } ( t ) = 2 \mathbf { i } - ( 3 t + 1 ) \mathbf { j } + 4 t \mathbf { k }$ and $\mathbf { w } ( t ) = 4 t \mathbf { i } + 5 \mathbf { j } - t ^ { 2 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

A) $\left( 9 t ^ { 2 } - 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$
B) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } - 36 t \mathbf { j } - ( 4 + 24 t ) \mathbf { k }$
C) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$
D) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } + 36 t \mathbf { j } - ( 4 + 24 t ) \mathbf { k }$
E) $\left( 9 t ^ { 2 } + 2 t - 20 \right) \mathbf { i } - 36 t \mathbf { j } + ( 4 + 24 t ) \mathbf { k }$

Accepted Answer

To find $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$, we use the product rule for differentiation, which in the context of vector cross products states that $\frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \frac{d\mathbf{u}}{dt} \times \mathbf{v} + \mathbf{u} \times \frac{d\mathbf{v}}{dt}$. Applying this rule to $\mathbf{v}(t)$ and $\mathbf{w}(t)$, and after computing the cross products and their derivatives, we obtain the expression corresponding to choice C.

Question 19

Let $\mathbf { v } ( t ) = 5 t \mathbf { i } - 2 t ^ { 2 } \mathbf { j } - \mathbf { k }$ and $\mathbf { w } ( t ) = - 4 t \mathbf { i } + t ^ { 2 } \mathbf { j } - 3 \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

A) $- 14 t \mathbf { i } + 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$
B) $14 t \mathbf { i } - 19 \mathbf { j } + 9 t ^ { 2 } \mathbf { k }$
C) $14 t \mathbf { i } + 19 \mathbf { j } + 9 t ^ { 2 } \mathbf { k }$
D) $14 t \mathbf { i } - 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$
E) $14 t \mathbf { i } + 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$

Accepted Answer

The derivative of the cross product $\mathbf{v}(t) \times \mathbf{w}(t)$ is calculated using the product rule for derivatives and the properties of cross products. The correct derivative simplifies to $14 t \mathbf{i} + 19 \mathbf{j} - 9 t^2 \mathbf{k}$.

Question 20

Let $\mathbf { v } ( t ) = 2 \mathbf { i } - 3 t \mathbf { j } + t ^ { 2 } \mathbf { k }$ and $\mathbf { w } ( t ) = - \mathbf { i } + 3 t \mathbf { j } + 5 t ^ { 2 } \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

A) $- 54 t ^ { 2 } \mathbf { i } + 22 t \mathbf { j } + 3 \mathbf { k }$
B) $- 54 t ^ { 2 } \mathbf { i } - 22 t \mathbf { j } + 3 \mathbf { k }$
C) $- 54 t ^ { 2 } \mathbf { i } - 22 t \mathbf { j } - 3 \mathbf { k }$
D) $- 54 t ^ { 2 } \mathbf { i } + 22 t \mathbf { j } - 3 \mathbf { k }$
E) $54 t ^ { 2 } \mathbf { i } - 22 t \mathbf { j } + 3 \mathbf { k }$

Accepted Answer

To find $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$, we can use the product rule for differentiation, which in the context of vector cross products states that $\frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \frac{d\mathbf{u}}{dt} \times \mathbf{v} + \mathbf{u} \times \frac{d\mathbf{v}}{dt}$. First, we differentiate $\mathbf{v}(t)$ and $\mathbf{w}(t)$ with respect to $t$, then compute the cross products and add them. The correct answer involves calculating these steps accurately, leading to the vector $- 54 t ^ { 2 } \mathbf { i } - 22 t \mathbf { j } + 3 \mathbf { k }$.

Quiz 12: Vector Functions

The domain of the vector function v(t) =eti−1−t2j+8k\mathbf { v } ( t ) = e ^ { t } \mathbf { i } - \sqrt { 1 - t ^ { 2 } } \mathbf { j } + 8 \mathbf { k }v(t) =eti−1−t2​j+8k is

The domain of the vector function v(t) =5t2i−ln⁡(3t−4) j+sin⁡tk\mathbf { v } ( t ) = 5 t ^ { 2 } \mathbf { i } - \ln ( 3 t - 4 ) \mathbf { j } + \sin t \mathbf { k }v(t) =5t2i−ln(3t−4) j+sintk is

The domain of the vector function v(t) =3t2−4i−sin⁡(2t) j+1tk\mathbf { v } ( t ) = \frac { 3 } { \sqrt { t ^ { 2 } - 4 } } \mathbf { i } - \sin ( 2 t ) \mathbf { j } + \frac { 1 } { t } \mathbf { k }v(t) =t2−4​3​i−sin(2t) j+t1​k is

The domain of the vector function v(t) =2ti−5t2j+log⁡2(t−4) k\mathbf { v } ( t ) = 2 ^ { t } \mathbf { i } - 5 t ^ { 2 } \mathbf { j } + \log _ { 2 } ( t - 4 ) \mathbf { k }v(t) =2ti−5t2j+log2​(t−4) k is

Let r(t) =cos⁡(2t−1) i−t2−1j+11+tk. Then \mathbf { r } ( t ) = \cos ( 2 t - 1 ) \mathbf { i } - \sqrt { t ^ { 2 } - 1 } \mathbf { j } + \frac { 1 } { 1 + t } \mathbf { k } \text {. Then }r(t) =cos(2t−1) i−t2−1​j+1+t1​k. Then is

Let r(t) =ecos⁡ti+tan⁡tj+cos⁡(5−t) k\mathbf { r } ( t ) = e ^ { \cos t } \mathbf { i } + \tan t \mathbf { j } + \cos ( 5 - t ) \mathbf { k }r(t) =ecosti+tantj+cos(5−t) k Then r′(t) \mathbf { r } ^ { \prime } ( t ) r′(t) is

Let r(t) =−sin⁡ti+cos⁡tj−e−tk\mathbf { r } ( t ) = - \sin t \mathbf { i } + \cos t \mathbf { j } - e ^ { - t } \mathbf { k }r(t) =−sinti+costj−e−tk Then r′(t) \mathbf { r } ^ { \prime } ( t ) r′(t) is

Let r(t) =esin⁡ti+ecos⁡tj−sec⁡tk\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + e ^ { \cos t } \mathbf { j } - \sec t \mathbf { k }r(t) =esinti+ecostj−sectk Then r′(t) \mathbf { r } ^ { \prime } ( t ) r′(t) is

Let r(t) =cos⁡(2t+1) i+t2+1j+t22k\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }r(t) =cos(2t+1) i+t2+1​j+2t2​k Then r′′(t) \mathbf { r } ^ { \prime \prime } ( t ) r′′(t) is

Let r(t) =esin⁡ti+cos⁡tj+ln⁡(5−t) k\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + \cos t \mathbf { j } + \ln ( 5 - t ) \mathbf { k }r(t) =esinti+costj+ln(5−t) k Then r′′(t) \mathbf { r } ^ { \prime \prime } ( t ) r′′(t) is

Let r(t) =−cos⁡ti+sin⁡tj−e−tk\mathbf { r } ( t ) = - \cos t \mathbf { i } + \sin t \mathbf { j } - e ^ { - t } \mathbf { k }r(t) =−costi+sintj−e−tk Then r′′(t) \mathbf { r } ^ { \prime \prime } ( t ) r′′(t) is

Let r(t) =esin⁡ti−ecos⁡tj+ln⁡tk\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } - e ^ { \cos t } \mathbf { j } + \ln t \mathbf { k }r(t) =esinti−ecostj+lntk Then r′′(t) \mathbf { r } ^ { \prime \prime } ( t ) r′′(t) is

The domain of the vector function $\mathbf { v } ( t ) = e ^ { t } \mathbf { i } - \sqrt { 1 - t ^ { 2 } } \mathbf { j } + 8 \mathbf { k }$ is

The domain of the vector function $\mathbf { v } ( t ) = 5 t ^ { 2 } \mathbf { i } - \ln ( 3 t - 4 ) \mathbf { j } + \sin t \mathbf { k }$ is

The domain of the vector function $\mathbf { v } ( t ) = \frac { 3 } { \sqrt { t ^ { 2 } - 4 } } \mathbf { i } - \sin ( 2 t ) \mathbf { j } + \frac { 1 } { t } \mathbf { k }$ is

The domain of the vector function $\mathbf { v } ( t ) = 2 ^ { t } \mathbf { i } - 5 t ^ { 2 } \mathbf { j } + \log _ { 2 } ( t - 4 ) \mathbf { k }$ is

Let $\mathbf { r } ( t ) = \cos ( 2 t - 1 ) \mathbf { i } - \sqrt { t ^ { 2 } - 1 } \mathbf { j } + \frac { 1 } { 1 + t } \mathbf { k } \text {. Then }$ is

Let $\mathbf { r } ( t ) = e ^ { \cos t } \mathbf { i } + \tan t \mathbf { j } + \cos ( 5 - t ) \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( t )$ is

Let $\mathbf { r } ( t ) = - \sin t \mathbf { i } + \cos t \mathbf { j } - e ^ { - t } \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( t )$ is

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + e ^ { \cos t } \mathbf { j } - \sec t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } ( t )$ is

Let $\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + \cos t \mathbf { j } + \ln ( 5 - t ) \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

Let $\mathbf { r } ( t ) = - \cos t \mathbf { i } + \sin t \mathbf { j } - e ^ { - t } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

Let $\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } - e ^ { \cos t } \mathbf { j } + \ln t \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is