| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | No, the given vector function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] does not parametrize a line. A line in 3D space is represented by a linear function of the parameter t. However, the given function r(t) contains terms with t raised to the power of 3 and 2, which means it is a higher-degree polynomial function. Therefore, it represents a curve in 3D space, not a line. To parametrize a line, the vector function should be of the form r(t) = p + td, where p is a point on the line and d is the direction vector of the line. In this case, the components of r(t) should be linear functions of t, not higher-degree polynomials. |
False |
| deepseekr1 | To determine if the parametric curve \( \mathbf{r}(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) parametrizes a line, we need to check if the direction vector is constant. 1. **Compute the velocity vector (first derivative):** \[ \mathbf{r}'(t) = \left[ \frac{d}{dt}(8 - 4t^3), \frac{d}{dt}(2 + 5t^2), \frac{d}{dt}(9t^3) \right] = [-12t^2, 10t, 27t^2] \] The velocity vector is not constant as it depends on \( t \). 2. **Check if the velocity vectors at different points are scalar multiples:** - At \( t = 0 \): \( \mathbf{r}(0) = [8, 2, 0] \) - At \( t = 1 \): \( \mathbf{r}(1) = [4, 7, 9] \) - Direction vector from \( t = 0 \) to \( t = 1 \): \( [-4, 5, 9] \) - At \( t = 2 \): \( \mathbf{r}(2) = [-24, 22, 72] \) - Direction vector from \( t = 1 \) to \( t = 2 \): \( [-28, 15, 63] \) - These vectors \( [-4, 5, 9] \) and \( [-28, 15, 63] \) are not scalar multiples, indicating the direction changes. 3. **Compute the acceleration vector (second derivative):** \[ \mathbf{r}''(t) = \left[ \frac{d}{dt}(-12t^2), \frac{d}{dt}(10t), \frac{d}{dt}(27t^2) \right] = [-24t, 10, 54t] \] The acceleration vector is non-zero and not collinear with the velocity vector, indicating the path is curved. 4. **Check cross product of velocity and acceleration:** - At \( t = 1 \): \( \mathbf{v} = [-12, 10, 27] \) and \( \mathbf{a} = [-24, 10, 54] \) - Cross product \( \mathbf{v} \times \mathbf{a} = [270, 0, 120] \neq \mathbf{0} \), showing they are not collinear. 5. **Check linear relationships:** - The curve lies on the plane \( 9x + 4z = 72 \), but \( y \) is non-linearly related to \( x \) and \( z \). Since the velocity vector is not constant, direction vectors are not scalar multiples, acceleration is not collinear with velocity, and relationships between coordinates are non-linear, the curve is not a line. \[ \boxed{\text{No}} \] |
False |
| gpto1 | To determine whether the vector function \( \mathbf{r}(t) = [8 - 4t^3, \ 2 + 5t^2, \ 9t^3] \) parametrizes a line, we need to analyze the path traced by \( \mathbf{r}(t) \) as \( t \) varies. A **line in three-dimensional space** can be represented parametrically by: \[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} \] where \( \mathbf{r}_0 \) is a fixed point on the line, and \( \mathbf{v} \) is a constant direction vector. In this representation, each component of \( \mathbf{r}(t) \) is a linear function of \( t \). Given that the components of \( \mathbf{r}(t) \) involve \( t^2 \) and \( t^3 \), they are not linear in \( t \). However, the nonlinearity in the parameter \( t \) does not immediately imply that the path is not a straight line. The function could still trace a straight line if the points \( \mathbf{r}(t) \) for different values of \( t \) lie along a straight line in space. **Approach to Determine if \( \mathbf{r}(t) \) Parametrizes a Line:** 1. **Analyze the Derivative \( \mathbf{r}'(t) \):** Compute the derivative of \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{r}'(t) = \left[ \frac{d}{dt}(8 - 4t^3),\ \frac{d}{dt}(2 + 5t^2),\ \frac{d}{dt}(9t^3) \right] = [-12t^2,\ 10t,\ 27t^2] \] The derivative \( \mathbf{r}'(t) \) represents the velocity vector at each point along the path. For a straight line, \( \mathbf{r}'(t) \) should be a constant vector (direction does not change with \( t \)). However, \( \mathbf{r}'(t) \) contains terms with \( t \), indicating that the direction of the velocity vector changes with \( t \). This suggests that the path is not a straight line. 2. **Attempt to Find a Linear Relationship Between \( x \), \( y \), and \( z \):** **From the given components:** \[ x(t) = 8 - 4t^3 \\ y(t) = 2 + 5t^2 \\ z(t) = 9t^3 \] **Relate \( x \) and \( z \):** From \( x(t) \) and \( z(t) \): \[ t^3 = \frac{8 - x}{4} \quad \text{and} \quad t^3 = \frac{z}{9} \] Equate the expressions for \( t^3 \): \[ \frac{8 - x}{4} = \frac{z}{9} \] Cross-multiplying: \[ 9(8 - x) = 4z \] Which simplifies to: \[ 9x + 4z = 72 \quad \text{or} \quad 9x + 4z - 72 = 0 \] This is a linear equation in \( x \) and \( z \). **Relate \( y \) and \( t^2 \):** From \( y(t) \): \[ y = 2 + 5t^2 \\ t^2 = \frac{y - 2}{5} \] **Attempt to relate \( x \) and \( y \):** Substitute \( t^3 \) in terms of \( y \): \[ t^3 = t \cdot t^2 = t \cdot \left( \frac{y - 2}{5} \right) \] But \( t \) is not directly expressed in terms of \( y \), and this introduces complications because \( t \) can be both positive and negative. **Conclusion from Relations:** While we obtained a linear equation \( 9x + 4z = 72 \), there is no consistent linear relationship involving all three variables \( x \), \( y \), and \( z \). Specifically, we cannot eliminate \( t \) to find a single linear equation that all the points satisfy. 3. **Test Specific Points Along \( \mathbf{r}(t) \):** **Calculate Points for Specific Values of \( t \):** - **At \( t = 0 \):** \[ x(0) = 8,\quad y(0) = 2,\quad z(0) = 0 \] Point \( A \): \( (8, 2, 0) \) - **At \( t = 1 \):** \[ x(1) = 8 - 4(1)^3 = 4, \quad y(1) = 2 + 5(1)^2 = 7, \quad z(1) = 9(1)^3 = 9 \] Point \( B \): \( (4, 7, 9) \) - **At \( t = 2 \):** \[ x(2) = 8 - 4(8) = -24, \quad y(2) = 2 + 5(4) = 22, \quad z(2) = 9(8) = 72 \] Point \( C \): \( (-24, 22, 72) \) **Compute Vectors Between Points:** - Vector \( \overrightarrow{AB} = \mathbf{B} - \mathbf{A} = (4 - 8,\ 7 - 2,\ 9 - 0) = (-4,\ 5,\ 9) \) - Vector \( \overrightarrow{AC} = \mathbf{C} - \mathbf{A} = (-24 - 8,\ 22 - 2,\ 72 - 0) = (-32,\ 20,\ 72) \) **Check for Collinearity:** For points \( A \), \( B \), and \( C \) to lie on a straight line, vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) must be scalar multiples of each other (i.e., parallel). Let's compare the components: - Ratio of \( x \)-components: \[ \frac{-32}{-4} = 8 \] - Ratio of \( y \)-components: \[ \frac{20}{5} = 4 \] - Ratio of \( z \)-components: \[ \frac{72}{9} = 8 \] The ratios are not equal (\( 8 \), \( 4 \), \( 8 \)), indicating that \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are not scalar multiples. Therefore, points \( A \), \( B \), and \( C \) do not lie along the same straight line. 4. **Compute the Cross Product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):** The cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) will be zero if the vectors are parallel (i.e., if the points lie along a straight line). \[ \overrightarrow{AB} = (-4,\ 5,\ 9) \\ \overrightarrow{AC} = (-32,\ 20,\ 72) \] **Compute the Cross Product:** \[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 5 & 9 \\ -32 & 20 & 72 \\ \end{vmatrix} \] - **\( \mathbf{i} \)-component:** \[ (5)(72) - (9)(20) = 360 - 180 = 180 \] - **\( \mathbf{j} \)-component:** \[ -\left[ (-4)(72) - (9)(-32) \right] = -\left( -288 + 288 \right) = 0 \] - **\( \mathbf{k} \)-component:** \[ (-4)(20) - (5)(-32) = -80 + 160 = 80 \] Thus, the cross product is: \[ \overrightarrow{AB} \times \overrightarrow{AC} = (180,\ 0,\ 80) \] Since the cross product is not the zero vector, \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are not parallel, confirming that the points do not lie on a straight line. **Conclusion:** The vector function \( \mathbf{r}(t) = [8 - 4t^3,\ 2 + 5t^2,\ 9t^3] \) does **not** parametrize a line. The changing direction of the derivative \( \mathbf{r}'(t) \), the lack of a consistent linear relationship among \( x \), \( y \), and \( z \), and the non-collinearity of points along \( \mathbf{r}(t) \) all indicate that the path traced by \( \mathbf{r}(t) \) is not a straight line. **Answer:** No; since \( \mathbf{r}(t) \) does not trace a straight line, it does not parametrize a line. |
False |
| GPT-4o | To determine whether the given vector function \( \mathbf{r}(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) parametrizes a line, we need to examine the nature of the function. A line in space can be parametrized by a linear function of the form: \[ \mathbf{r}(t) = \mathbf{r_0} + t \mathbf{d} \] where \(\mathbf{r_0}\) is a point on the line and \(\mathbf{d}\) is a direction vector. We can look at each component of \(\mathbf{r}(t)\): 1. \(x(t) = 8 - 4t^3\) 2. \(y(t) = 2 + 5t^2\) 3. \(z(t) = 9t^3\) For \(\mathbf{r}(t)\) to parametrize a line, each component function must be a linear function of \(t\). - \(x(t) = 8 - 4t^3\) is not linear because it contains a \(t^3\) term. - \(y(t) = 2 + 5t^2\) is not linear because it contains a \(t^2\) term. - \(z(t) = 9t^3\) is not linear because it contains a \(t^3\) term. Since none of the component functions \(x(t)\), \(y(t)\), and \(z(t)\) are linear in \(t\), the given vector function \(\mathbf{r}(t)\) does not parametrize a line. Therefore, the solution is: \[ \mathbf{r}(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \text{ does not parametrize a line.} \] |
False |
| llama3.1-70B-Instruct | To determine if the given vector function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] parametrizes a line, we need to check if the function satisfies the properties of a line. A line in 3D space can be represented by a vector function of the form: r(t) = a + tb where a is a point on the line, b is a direction vector of the line, and t is a parameter. Comparing the given function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] with the general form of a line, we can see that: * The first component 8 - 4t^3 does not have a linear dependence on t, as it involves a cubic term. * The second component 2 + 5t^2 does not have a linear dependence on t, as it involves a quadratic term. * The third component 9t^3 does not have a linear dependence on t, as it involves a cubic term. Since none of the components of the given function have a linear dependence on t, the function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] does not satisfy the properties of a line. Therefore, the given function does not parametrize a line. |
False |
| llama3.1-405B-Instruct | To determine if the vector function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] parametrizes a line, we need to check if it can be written in the form r(t) = a + bt, where a and b are constant vectors. Let's try to rewrite r(t) in this form: r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] We can see that the first component, 8 - 4t^3, is not a linear function of t, so it cannot be written in the form a + bt. Therefore, r(t) does not parametrize a line. Alternatively, we can also check if the vector function r(t) satisfies the condition that r(t) - r(0) is parallel to r(1) - r(0) for all t. If this condition is satisfied, then r(t) parametrizes a line. Let's calculate r(0) and r(1): r(0) = [8, 2, 0] r(1) = [4, 7, 9] Now, let's calculate r(t) - r(0) and r(1) - r(0): r(t) - r(0) = [-4t^3, 5t^2, 9t^3] r(1) - r(0) = [-4, 5, 9] We can see that r(t) - r(0) is not parallel to r(1) - r(0) for all t, since the first component of r(t) - r(0) is a cubic function of t, while the first component of r(1) - r(0) is a constant. Therefore, r(t) does not parametrize a line. |
False |
| DeepSeek-Coder-V2 | To determine if the given vector function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] parametrizes a line, we need to check if the direction of the line remains constant as t varies. A line can be parametrized if the direction vector (the difference between any two points on the line) is constant. Let's consider two points on the curve, r(t1) and r(t2), where t1 and t2 are any two values of t. The direction vector between these two points is given by: r(t2) - r(t1) = [(8 - 4t2^3) - (8 - 4t1^3), (2 + 5t2^2) - (2 + 5t1^2), (9t2^3) - (9t1^3)] Simplifying the components, we get: r(t2) - r(t1) = [-4t2^3 + 4t1^3, 5t2^2 - 5t1^2, 9t2^3 - 9t1^3] Now, let's factor out the common terms: r(t2) - r(t1) = [-4(t2^3 - t1^3), 5(t2^2 - t1^2), 9(t2^3 - t1^3)] We can further factor the expressions inside the parentheses: r(t2) - r(t1) = [-4(t2 - t1)(t2^2 + t2t1 + t1^2), 5(t2 - t1)(t2 + t1), 9(t2 - t1)(t2^2 + t2t1 + t1^2)] Notice that the direction vector r(t2) - r(t1) is not constant, as it depends on the values of t1 and t2. The direction vector changes as t varies, which means the curve described by r(t) is not a line. Therefore, the given vector function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] does not parametrize a line. |
False |
| Qwen2-72B-Instruct | No, the given vector function r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] does not parametrize a line. A line in 3D space is represented by a linear function of the parameter t. However, the given function r(t) contains terms with t raised to the power of 3 and 2, which means it is a higher-degree polynomial function. Therefore, it represents a curve in 3D space, not a line. To parametrize a line, the vector function should be of the form r(t) = p + td, where p is a point on the line and d is the direction vector of the line. In this case, the components of r(t) should be linear functions of t, not higher-degree polynomials. |
False |
| baichuan4 | To determine if the vector-valued function \(r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3]\) parametrizes a line, we need to check if it satisfies the condition for a line in three-dimensional space. A vector-valued function parametrizes a line if it can be written in the form: \[r(t) = r_0 + tv\] where \(r_0\) is a point on the line (a constant vector), and \(v\) is a constant vector representing the direction of the line. Let's examine the given function: \[r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3]\] We can rewrite this as: \[r(t) = [8, 2, 0] + t[-4t^2, 5t, 9t^2]\] Here, \([8, 2, 0]\) is a constant vector (which could represent a point on the line), but the direction vector \([-4t^2, 5t, 9t^2]\) is not constant because it contains variable \(t\). Therefore, this function does not satisfy the condition for a line. So, the answer to the problem is: No, the given function \(r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3]\) does not parametrize a line because the direction vector is not constant. |
False |
| doubao | For a vector function to parametrize a line, all its components must be of the form at + b (where a and b are constants). Looking at the given r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3], the first component 8 - 4t^3 is not of the form at + b, nor is the second component 2 + 5t^2. Only the third component 9t^3 is of a form similar to at (with a = 9 and b = 0), but since not all components have this form, r(t) does not parametrize a line. |
False |
| minimax | To determine whether the given vector function \( r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) parametrizes a line, we need to check if it represents a straight path in three-dimensional space. A line in three-dimensional space can be represented by a vector function of the form: \[ r(t) = r_0 + t\mathbf{d} \] where \( r_0 \) is a position vector to a point on the line (the initial point), \( \mathbf{d} \) is a constant direction vector, and \( t \) is a scalar parameter. Let's analyze the given vector function \( r(t) \): \[ r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \] We can see that the components of \( r(t) \) are not linear functions of \( t \). Specifically, the first component is \( 8 - 4t^3 \), the second component is \( 2 + 5t^2 \), and the third component is \( 9t^3 \). The presence of \( t^2 \) and \( t^3 \) terms indicates that the path is not a straight line because the position vector changes direction as \( t \) changes, due to the non-linear terms. For a line, we would expect each component to be a linear function of \( t \), such as \( at + b \), where \( a \) and \( b \) are constants. In this case, the components are not linear functions of \( t \), but rather polynomial functions of \( t \) with different powers. Therefore, the given vector function \( r(t) \) does not parametrize a line; it parametrizes a curve in three-dimensional space. |
False |
| kimi | To determine whether the vector function \( r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) parametrizes a line, we need to check if the cross product of the derivatives of the individual components with respect to \( t \) is zero. If the cross product is zero, then the direction vectors are linearly dependent, which means the curve lies in a plane, and thus it cannot be a line. Let's find the derivatives of each component with respect to \( t \): \[ r_x'(t) = \frac{d}{dt}(8 - 4t^3) = -12t^2 \] \[ r_y'(t) = \frac{d}{dt}(2 + 5t^2) = 10t \] \[ r_z'(t) = \frac{d}{dt}(9t^3) = 27t^2 \] Now, let's compute the cross product of these derivatives: \[ \vec{r_x'}(t) \times \vec{r_y'}(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -12t^2 & 10t & 0 \\ \end{vmatrix} \] Expanding this determinant, we get: \[ \vec{r_x'}(t) \times \vec{r_y'}(t) = \mathbf{i}(10t \cdot 0 - 0 \cdot 27t^2) - \mathbf{j}(-12t^2 \cdot 0 - 0 \cdot -12t^2) + \mathbf{k}(-12t^2 \cdot 10t - 10t \cdot -12t^2) \] \[ = \mathbf{i}(0) - \mathbf{j}(0) + \mathbf{k}(-120t^3 + 120t^3) \] \[ = \mathbf{0} \] Since the cross product is the zero vector, it means that the direction vectors \( \vec{r_x'}(t) \) and \( \vec{r_y'}(t) \) are linearly dependent, and thus the curve is not a line but a curve that lies in a plane. Therefore, the vector function \( r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) does not parametrize a line. |
False |
| wenxinyiyan | To determine if r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] parametrizes a line, we need to check if the velocity vector is constant. The velocity vector is the derivative of the position vector with respect to time, t. Taking the derivative of each component of r(t), we get: r'(t) = d/dt [8 - 4t^3, 2 + 5t^2, 9t^3] = [-12t^2, 10t, 27t^2] If r'(t) is constant, then each component must be a constant function of t. However, we see that each component of r'(t) is a function of t, and not constant. Therefore, r(t) does not parametrize a line. |
False |
| zhipu | To determine whether the vector function \( r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) parametrizes a line, we need to check if it can be expressed in the form of a linear function of \( t \). A line in three-dimensional space can be parametrized as: \[ r(t) = \vec{a} + t\vec{d} \] where \( \vec{a} \) is a position vector (a point on the line) and \( \vec{d} \) is a direction vector (indicating the direction of the line), and \( t \) is a real number parameter. The components of \( r(t) \) should all be linear functions of \( t \) (that is, of the form \( at + b \), where \( a \) and \( b \) are constants) if \( r(t) \) is to parametrize a line. Let's examine each component of \( r(t) \): 1. The x-component is \( 8 - 4t^3 \). This is a cubic function of \( t \), not a linear function. 2. The y-component is \( 2 + 5t^2 \). This is a quadratic function of \( t \), not a linear function. 3. The z-component is \( 9t^3 \). This is also a cubic function of \( t \), not a linear function. Since none of the components of \( r(t) \) are linear functions of \( t \), the vector function \( r(t) \) does not parametrize a line. Therefore, the solution to the problem is: **No, \( r(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) does not parametrize a line because its components are not linear functions of the parameter \( t \).** |
False |
| gpto1 | To determine whether the vector function \( \mathbf{r}(t) = [8 - 4t^3, \ 2 + 5t^2, \ 9t^3] \) parametrizes a line, we need to analyze the path traced by \( \mathbf{r}(t) \) as \( t \) varies. A **line in three-dimensional space** can be represented parametrically by: \[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} \] where \( \mathbf{r}_0 \) is a fixed point on the line, and \( \mathbf{v} \) is a constant direction vector. In this representation, each component of \( \mathbf{r}(t) \) is a linear function of \( t \). Given that the components of \( \mathbf{r}(t) \) involve \( t^2 \) and \( t^3 \), they are not linear in \( t \). However, the nonlinearity in the parameter \( t \) does not immediately imply that the path is not a straight line. The function could still trace a straight line if the points \( \mathbf{r}(t) \) for different values of \( t \) lie along a straight line in space. **Approach to Determine if \( \mathbf{r}(t) \) Parametrizes a Line:** 1. **Analyze the Derivative \( \mathbf{r}'(t) \):** Compute the derivative of \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{r}'(t) = \left[ \frac{d}{dt}(8 - 4t^3),\ \frac{d}{dt}(2 + 5t^2),\ \frac{d}{dt}(9t^3) \right] = [-12t^2,\ 10t,\ 27t^2] \] The derivative \( \mathbf{r}'(t) \) represents the velocity vector at each point along the path. For a straight line, \( \mathbf{r}'(t) \) should be a constant vector (direction does not change with \( t \)). However, \( \mathbf{r}'(t) \) contains terms with \( t \), indicating that the direction of the velocity vector changes with \( t \). This suggests that the path is not a straight line. 2. **Attempt to Find a Linear Relationship Between \( x \), \( y \), and \( z \):** **From the given components:** \[ x(t) = 8 - 4t^3 \\ y(t) = 2 + 5t^2 \\ z(t) = 9t^3 \] **Relate \( x \) and \( z \):** From \( x(t) \) and \( z(t) \): \[ t^3 = \frac{8 - x}{4} \quad \text{and} \quad t^3 = \frac{z}{9} \] Equate the expressions for \( t^3 \): \[ \frac{8 - x}{4} = \frac{z}{9} \] Cross-multiplying: \[ 9(8 - x) = 4z \] Which simplifies to: \[ 9x + 4z = 72 \quad \text{or} \quad 9x + 4z - 72 = 0 \] This is a linear equation in \( x \) and \( z \). **Relate \( y \) and \( t^2 \):** From \( y(t) \): \[ y = 2 + 5t^2 \\ t^2 = \frac{y - 2}{5} \] **Attempt to relate \( x \) and \( y \):** Substitute \( t^3 \) in terms of \( y \): \[ t^3 = t \cdot t^2 = t \cdot \left( \frac{y - 2}{5} \right) \] But \( t \) is not directly expressed in terms of \( y \), and this introduces complications because \( t \) can be both positive and negative. **Conclusion from Relations:** While we obtained a linear equation \( 9x + 4z = 72 \), there is no consistent linear relationship involving all three variables \( x \), \( y \), and \( z \). Specifically, we cannot eliminate \( t \) to find a single linear equation that all the points satisfy. 3. **Test Specific Points Along \( \mathbf{r}(t) \):** **Calculate Points for Specific Values of \( t \):** - **At \( t = 0 \):** \[ x(0) = 8,\quad y(0) = 2,\quad z(0) = 0 \] Point \( A \): \( (8, 2, 0) \) - **At \( t = 1 \):** \[ x(1) = 8 - 4(1)^3 = 4, \quad y(1) = 2 + 5(1)^2 = 7, \quad z(1) = 9(1)^3 = 9 \] Point \( B \): \( (4, 7, 9) \) - **At \( t = 2 \):** \[ x(2) = 8 - 4(8) = -24, \quad y(2) = 2 + 5(4) = 22, \quad z(2) = 9(8) = 72 \] Point \( C \): \( (-24, 22, 72) \) **Compute Vectors Between Points:** - Vector \( \overrightarrow{AB} = \mathbf{B} - \mathbf{A} = (4 - 8,\ 7 - 2,\ 9 - 0) = (-4,\ 5,\ 9) \) - Vector \( \overrightarrow{AC} = \mathbf{C} - \mathbf{A} = (-24 - 8,\ 22 - 2,\ 72 - 0) = (-32,\ 20,\ 72) \) **Check for Collinearity:** For points \( A \), \( B \), and \( C \) to lie on a straight line, vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) must be scalar multiples of each other (i.e., parallel). Let's compare the components: - Ratio of \( x \)-components: \[ \frac{-32}{-4} = 8 \] - Ratio of \( y \)-components: \[ \frac{20}{5} = 4 \] - Ratio of \( z \)-components: \[ \frac{72}{9} = 8 \] The ratios are not equal (\( 8 \), \( 4 \), \( 8 \)), indicating that \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are not scalar multiples. Therefore, points \( A \), \( B \), and \( C \) do not lie along the same straight line. 4. **Compute the Cross Product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):** The cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) will be zero if the vectors are parallel (i.e., if the points lie along a straight line). \[ \overrightarrow{AB} = (-4,\ 5,\ 9) \\ \overrightarrow{AC} = (-32,\ 20,\ 72) \] **Compute the Cross Product:** \[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 5 & 9 \\ -32 & 20 & 72 \\ \end{vmatrix} \] - **\( \mathbf{i} \)-component:** \[ (5)(72) - (9)(20) = 360 - 180 = 180 \] - **\( \mathbf{j} \)-component:** \[ -\left[ (-4)(72) - (9)(-32) \right] = -\left( -288 + 288 \right) = 0 \] - **\( \mathbf{k} \)-component:** \[ (-4)(20) - (5)(-32) = -80 + 160 = 80 \] Thus, the cross product is: \[ \overrightarrow{AB} \times \overrightarrow{AC} = (180,\ 0,\ 80) \] Since the cross product is not the zero vector, \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are not parallel, confirming that the points do not lie on a straight line. **Conclusion:** The vector function \( \mathbf{r}(t) = [8 - 4t^3,\ 2 + 5t^2,\ 9t^3] \) does **not** parametrize a line. The changing direction of the derivative \( \mathbf{r}'(t) \), the lack of a consistent linear relationship among \( x \), \( y \), and \( z \), and the non-collinearity of points along \( \mathbf{r}(t) \) all indicate that the path traced by \( \mathbf{r}(t) \) is not a straight line. **Answer:** No; since \( \mathbf{r}(t) \) does not trace a straight line, it does not parametrize a line. |
False |
| deepseekv3 | To determine whether the vector function \( \mathbf{r}(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) parametrizes a line, we need to inspect the components of the vector function. ### Key Insight: A vector function parametrizes a line **if and only if** all its components are linear functions of the parameter \( t \). This means each component must be of the form \( at + b \), where \( a \) and \( b \) are constants. ### Analysis of \( \mathbf{r}(t) \): 1. The first component is \( 8 - 4t^3 \). This is a **cubic** function of \( t \), not linear. 2. The second component is \( 2 + 5t^2 \). This is a **quadratic** function of \( t \), not linear. 3. The third component is \( 9t^3 \). This is a **cubic** function of \( t \), not linear. ### Conclusion: Since none of the components of \( \mathbf{r}(t) \) are linear functions of \( t \), the vector function \( \mathbf{r}(t) \) does **not** parametrize a line. ### Final Answer: No, \( \mathbf{r}(t) = [8 - 4t^3, 2 + 5t^2, 9t^3] \) does not parametrize a line because its components are not linear functions of \( t \). |
False |