Problem 3 For Exercises \(1-4,\) calculate... [FREE SOLUTION] (2024)

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When you face problems involving finding the tangent line to a curve at a specific point, it's crucial to understand the formula and the steps involved. The tangent line at any point is essentially a straight line that just 'touches' the curve at that point, moving in the same direction as the curve's slope at that point.

The general formula for the tangent line to a function \({\mathbf{f}(t)}\) at \(t = a\) is:
\((x, y, z) = \mathbf{f}(a) + t \cdot \mathbf{f'}(0)\).

Break down the formula:

• \mathbf{f}(a)\: This gives you the point on the curve where you want to find the tangent line.
• \mathbf{f'}(0)\: This is the derivative evaluated at \(t = a\). It gives the direction in which the tangent line points.

Always remember to simplify the equation of the tangent line at the end to make it easier to understand and graph.

Understanding how to find the derivative of vector functions is critical in vector calculus. A vector function consists of multiple component functions (e.g., \(\mathbf{f}(t) = (\text{cos} 2t, \text{sin} 2t, t)\)). To find its derivative, you need to handle each component individually.

For example, for the function \ \mathbf{f}(t) = (\cos 2t, \sin 2t, t) \ the derivatives are calculated as:

• \frac{d}{dt}(\cos 2t) = -2\sin 2t
• \frac{d}{dt}(\sin 2t) = 2\cos 2t
• \frac{d}{dt}(t) = 1

Hence, the derivative vector becomes:
\mathbf{f'}(t) = (-2 \sin 2t, 2 \cos 2t, 1) \.

Make sure to find the derivative of each component in the vector separately. Finally, combine these to form the derivative vector. This forms the basis for further calculations like finding tangent lines, understanding motion, and more.

Evaluating the derivatives of vector functions at specific points is a common step in many vector calculus problems. Once you've found the general derivative \mathbf{f'}(t), you often need to evaluate it at a specific point \(t = a\).

Using our example, we found that \mathbf{f'}(t) = (-2 \sin 2t, 2 \cos 2t, 1).
When we evaluate this at \(t = 0\), we get:

• \mathbf{f'}(0) = (-2 \sin (2 \cdot 0), 2 \cos (2 \cdot 0), 1) = (0, 2, 1)

Being thorough in evaluating the derivative is essential, as errors can lead to misinterpretations of tangent lines and other related computations. Make sure to methodically substitute \(t = a\) and simplify each term step-by-step. This approach ensures that you get accurate and reliable results.