Answer :
Sure! Let's find the parametric equations for the tangent line to the curve given by the parametric equations [tex]\( x = t^2 + 1 \)[/tex], [tex]\( y = 8\sqrt{t} \)[/tex], and [tex]\( z = e^{t^2 - t} \)[/tex] at the specified point [tex]\((2, 8, 1)\)[/tex].
1. Determine the value of [tex]\( t \)[/tex] at the point (2, 8, 1)
First, we need to find the value of [tex]\( t \)[/tex] when the curve passes through the point [tex]\((2, 8, 1)\)[/tex].
- For [tex]\( x \)[/tex]:
[tex]\[ x = t^2 + 1 \implies 2 = t^2 + 1 \implies t^2 = 1 \implies t = \pm 1 \][/tex]
- For [tex]\( y \)[/tex]:
[tex]\[ y = 8 \sqrt{t} \implies 8 = 8 \sqrt{t} \implies \sqrt{t} = 1 \implies t = 1 \][/tex]
- For [tex]\( z \)[/tex]:
[tex]\[ z = e^{t^2 - t} \implies 1 = e^{t^2 - t} \implies t^2 - t = 0 \implies t(t-1) = 0 \implies t = 0 \text{ or } t = 1 \][/tex]
Given that [tex]\( t = 1 \)[/tex] satisfies all the equations simultaneously, we have [tex]\( t = 1 \)[/tex].
2. Calculate derivatives at the point
Next, we'll find the derivatives of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] with respect to [tex]\( t \)[/tex] at [tex]\( t = 1 \)[/tex].
- For [tex]\( \frac{dx}{dt} \)[/tex]:
[tex]\[ x = t^2 + 1 \implies \frac{dx}{dt} = 2t \implies \frac{dx}{dt} \bigg|_{t=1} = 2 \cdot 1 = 2 \][/tex]
- For [tex]\( \frac{dy}{dt} \)[/tex]:
[tex]\[ y = 8 \sqrt{t} \implies \frac{dy}{dt} = 8 \cdot \frac{1}{2\sqrt{t}} = \frac{8}{2\sqrt{t}} = \frac{4}{\sqrt{t}} \implies \frac{dy}{dt} \bigg|_{t=1} = \frac{4}{\sqrt{1}} = 4 \][/tex]
- For [tex]\( \frac{dz}{dt} \)[/tex]:
[tex]\[ z = e^{t^2 - t} \implies \frac{dz}{dt} = e^{t^2 - t} \cdot \frac{d}{dt} (t^2 - t) = e^{t^2 - t} (2t-1) \implies \frac{dz}{dt} \bigg|_{t=1} = e^{1-1} \cdot (2 \cdot 1 - 1) = 1 \cdot 1 = 1 \][/tex]
3. Write the parametric equations for the tangent line
The parametric equations for the tangent line at [tex]\((2, 8, 1)\)[/tex] can be expressed as:
[tex]\[ \begin{cases} x = x_0 + \left(\frac{dx}{dt}\bigg|_{t=1}\right)(t) \\ y = y_0 + \left(\frac{dy}{dt}\bigg|_{t=1}\right)(t) \\ z = z_0 + \left(\frac{dz}{dt}\bigg|_{t=1}\right)(t) \end{cases} \][/tex]
By plugging in values [tex]\((x_0, y_0, z_0) = (2, 8, 1)\)[/tex], [tex]\(\frac{dx}{dt} = 2\)[/tex], [tex]\(\frac{dy}{dt} = 4\)[/tex], [tex]\(\frac{dz}{dt} = 1\)[/tex]:
[tex]\[ \begin{cases} x = 2 + 2t \\ y = 8 + 4t \\ z = 1 + t \end{cases} \][/tex]
Therefore, the parametric equations for the tangent line are:
[tex]\[ \boxed{ \begin{cases} x = 2 + 2t \\ y = 8 + 4t \\ z = 1 + t \end{cases} } \][/tex]
1. Determine the value of [tex]\( t \)[/tex] at the point (2, 8, 1)
First, we need to find the value of [tex]\( t \)[/tex] when the curve passes through the point [tex]\((2, 8, 1)\)[/tex].
- For [tex]\( x \)[/tex]:
[tex]\[ x = t^2 + 1 \implies 2 = t^2 + 1 \implies t^2 = 1 \implies t = \pm 1 \][/tex]
- For [tex]\( y \)[/tex]:
[tex]\[ y = 8 \sqrt{t} \implies 8 = 8 \sqrt{t} \implies \sqrt{t} = 1 \implies t = 1 \][/tex]
- For [tex]\( z \)[/tex]:
[tex]\[ z = e^{t^2 - t} \implies 1 = e^{t^2 - t} \implies t^2 - t = 0 \implies t(t-1) = 0 \implies t = 0 \text{ or } t = 1 \][/tex]
Given that [tex]\( t = 1 \)[/tex] satisfies all the equations simultaneously, we have [tex]\( t = 1 \)[/tex].
2. Calculate derivatives at the point
Next, we'll find the derivatives of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] with respect to [tex]\( t \)[/tex] at [tex]\( t = 1 \)[/tex].
- For [tex]\( \frac{dx}{dt} \)[/tex]:
[tex]\[ x = t^2 + 1 \implies \frac{dx}{dt} = 2t \implies \frac{dx}{dt} \bigg|_{t=1} = 2 \cdot 1 = 2 \][/tex]
- For [tex]\( \frac{dy}{dt} \)[/tex]:
[tex]\[ y = 8 \sqrt{t} \implies \frac{dy}{dt} = 8 \cdot \frac{1}{2\sqrt{t}} = \frac{8}{2\sqrt{t}} = \frac{4}{\sqrt{t}} \implies \frac{dy}{dt} \bigg|_{t=1} = \frac{4}{\sqrt{1}} = 4 \][/tex]
- For [tex]\( \frac{dz}{dt} \)[/tex]:
[tex]\[ z = e^{t^2 - t} \implies \frac{dz}{dt} = e^{t^2 - t} \cdot \frac{d}{dt} (t^2 - t) = e^{t^2 - t} (2t-1) \implies \frac{dz}{dt} \bigg|_{t=1} = e^{1-1} \cdot (2 \cdot 1 - 1) = 1 \cdot 1 = 1 \][/tex]
3. Write the parametric equations for the tangent line
The parametric equations for the tangent line at [tex]\((2, 8, 1)\)[/tex] can be expressed as:
[tex]\[ \begin{cases} x = x_0 + \left(\frac{dx}{dt}\bigg|_{t=1}\right)(t) \\ y = y_0 + \left(\frac{dy}{dt}\bigg|_{t=1}\right)(t) \\ z = z_0 + \left(\frac{dz}{dt}\bigg|_{t=1}\right)(t) \end{cases} \][/tex]
By plugging in values [tex]\((x_0, y_0, z_0) = (2, 8, 1)\)[/tex], [tex]\(\frac{dx}{dt} = 2\)[/tex], [tex]\(\frac{dy}{dt} = 4\)[/tex], [tex]\(\frac{dz}{dt} = 1\)[/tex]:
[tex]\[ \begin{cases} x = 2 + 2t \\ y = 8 + 4t \\ z = 1 + t \end{cases} \][/tex]
Therefore, the parametric equations for the tangent line are:
[tex]\[ \boxed{ \begin{cases} x = 2 + 2t \\ y = 8 + 4t \\ z = 1 + t \end{cases} } \][/tex]
