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Chapter 2: Problem 17

Determine the equation of the tangent line to the given path at the specified value of \(t.\) $$\left(\sin 3 t, \cos 3 t, 2 t^{5 / 2}\right) ; t=1$$

Short Answer

Expert verified
The tangent line is parameterized by: \( x = \sin(3) + 3t\cos(3), y = \cos(3) - 3t\sin(3), z = 2 + 5t \).

Step by step solution

01

Define the Parametric Equations

The given parametric equations are \( x = \sin(3t) \), \( y = \cos(3t) \), and \( z = 2t^{5/2} \). Identify these as the functions describing the path.
02

Find the Point at \(t = 1\)

Substitute \( t = 1 \) into the parametric equations to find the coordinates of the point. Thus, \( x(1) = \sin(3 \cdot 1) = \sin(3) \), \( y(1) = \cos(3 \cdot 1) = \cos(3) \), and \( z(1) = 2(1)^{5/2} = 2 \). Thus, the point is \((\sin(3), \cos(3), 2)\).
03

Find the Velocity Vector

Calculate the derivatives of the parametric equations with respect to \(t\) to get the components of the velocity vector.\[ \frac{dx}{dt} = 3\cos(3t) \], \[ \frac{dy}{dt} = -3\sin(3t) \], and \[ \frac{dz}{dt} = 5t^{3/2} \].
04

Evaluate the Velocity at \(t = 1\)

Substitute \( t = 1 \) into the derivatives to find the velocity vector at this instant.\[ \frac{dx}{dt}(1) = 3\cos(3) \], \[ \frac{dy}{dt}(1) = -3\sin(3) \], and \[ \frac{dz}{dt}(1) = 5(1)^{3/2} = 5 \]. The velocity vector is \((3\cos(3), -3\sin(3), 5)\).
05

Write the Equation of the Tangent Line

The equation of the tangent line is given by the point and the direction vector (velocity vector). Write the equation in parametric form: \( x = \sin(3) + 3t\cos(3) \), \( y = \cos(3) - 3t\sin(3) \), \( z = 2 + 5t \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations are a way of representing mathematical curves. Instead of expressing variables like x and y directly in terms of each other, each is expressed as a function of a third variable, usually t. This method is especially useful for plotting curves in multiple dimensions.
  • In the given problem, the parametric equations are: \( x = \sin(3t) \), \( y = \cos(3t) \), and \( z = 2t^{5/2} \).
  • Each equation describes how the x, y, and z positions change with t.
  • Visualize it like plotting the path of an airplane flying through 3D space.
By setting \( t = 1 \), we find the exact position of the object at that particular time on this path. This approach allows us to not only know the position but also to analyze the motion along the path.
Velocity Vector
The velocity vector provides insight into the speed and direction of an object at a particular instance on its path. It is derived by taking the derivatives of the parametric equations with respect to the variable t.
  • To find the velocity, calculate the derivatives: \( \frac{dx}{dt} = 3\cos(3t) \), \( \frac{dy}{dt} = -3\sin(3t) \), and \( \frac{dz}{dt} = 5t^{3/2} \).
  • These derivatives reflect how fast and in which direction the parameters x, y, and z change as t changes.
For \( t = 1 \), the velocity vector becomes \( (3\cos(3), -3\sin(3), 5) \). This vector points in the direction of the tangent line to the path at the point \( (\sin(3), \cos(3), 2) \). Understanding this vector helps determine the real-time movement and orientation of objects on their paths.
Calculus
Calculus is an essential mathematical tool for understanding change and motion. It enables us to determine slopes, curvatures, and the behavior of functions.

Derivatives

Derivatives play a crucial role. They help find the rate at which a function is changing. By calculating the derivative, we acquire the slope of a curve or the rate of change at any given point.
  • The derivative of a position function gives the velocity, which we used earlier to find the velocity vector.
  • This approach allows us to predict how an object moves along a curved path.

Tangent Lines

A tangent line represents the best linear approximation to a curve at a particular point.
  • By combining a point obtained from the parametric equations and the velocity vector, we define the tangent line.
  • This line is instrumental when analyzing real-world problems like motion trajectories and assessing instantaneous changes.
Understanding these concepts within calculus gives students powerful tools to tackle complex paths and motions analytically.

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Most popular questions from this chapter

Prove that if \(f: U \subset \mathbb{R}^{n} \rightarrow \mathbb{R}\) is differentiable at \(\mathbf{x}_{0} \in U,\) there is a neighborhood \(V\) of \(\mathbf{0} \in \mathbb{R}^{n}\) and a function \(R_{1}: V \rightarrow \mathbb{R}\) such that for all \(\mathbf{h} \in V,\) we have \(\mathbf{x}_{0}+\mathbf{h} \in U\) $$ f\left(\mathbf{x}_{0}+\mathbf{h}\right)=f\left(\mathbf{x}_{0}\right)+\left[\mathbf{D} f\left(\mathbf{x}_{0}\right)\right] \mathbf{h}+R_{1}(\mathbf{h}) $$ and $$ \frac{R_{1}(\mathbf{h})}{\|\mathbf{h}\|} \rightarrow 0 \quad \text { as } \quad \mathbf{h} \rightarrow \mathbf{0} $$

Use the chain rule and differentiation under the integral sign, namely, $$ \frac{d}{d x} \int_{a}^{b} f(x, y) d y=\int_{a}^{b} \frac{\partial f}{\partial x}(x, y) d y $$ to show that $$ \frac{d}{d x} \int_{0}^{x} f(x, y) d y=f(x, x)+\int_{0}^{x} \frac{\partial f}{\partial x}(x, y) d y $$

If \(z=f(x-y),\) use the chain rule to show that \(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=0\)

Suppose \(\mathbf{x}_{0} \in \mathbb{R}^{n}\) and \(0 \leq r_{1} < r_{2} .\) Show that there is a \(C^{1}\) function \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) such that \(f(\mathbf{x})=0\) for \(\left\|\mathbf{x}-\mathbf{x}_{0}\right\| \geq r_{2} ; 0 < f(\mathbf{x}) < 1\) for \(r_{1} < \left\|\mathbf{x}-\mathbf{x}_{0}\right\| < r_{2}\) and \(f(\mathbf{x})=1\) for \(\left\|\mathbf{x}-\mathbf{x}_{0}\right\| \leq r_{1} .\) [HINT: Apply a cubic polynomial with \(g\left(r_{1}^{2}\right)=1\) and \(g\left(r_{2}^{2}\right)=g^{\prime}\left(r_{2}^{2}\right)=\) \(\left.g^{\prime}\left(r_{1}^{2}\right)=0 \text { to }\left\|\mathbf{x}-\mathbf{x}_{0}\right\|^{2} \text { when } r_{1} < \left\|\mathbf{x}-\mathbf{x}_{0}\right\| < r_{2} .\right]\)

Describe the graph of each function by computing some level sets and sections. $$f: \mathbb{R}^{2} \rightarrow \mathbb{R},(x, y) \mapsto|y|$$
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