At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
