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Chapter 12: Problem 41

Evaluate the following limits. $$\lim _{t \rightarrow \pi / 2}\left(\cos 2 t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2 t}{\pi} \mathbf{k}\right)$$

Short Answer

Expert verified
$$ Answer: As t approaches 蟺/2, the limit of the vector-valued function is: $$(-1\mathbf{i} - 4\mathbf{j} + 1\mathbf{k})$$

Step by step solution

01

Evaluate the limit for the i-component

The i-component of the given vector function is cos(2t). To find its limit as t approaches 蟺/2, simply substitute 蟺/2 into the function: $$\lim_{t \rightarrow \pi/2} \cos(2t) = \cos(2(\pi/2)) = \cos(\pi)$$ Since cos(蟺) = -1, the limit for the i-component is: $$\lim_{t \rightarrow \pi/2}\cos(2t) \mathbf{i} = -1\mathbf{i}$$
02

Evaluate the limit for the j-component

The j-component of the given vector function is -4sin(t). To find its limit as t approaches 蟺/2, substitute 蟺/2 into the function: $$\lim_{t \rightarrow \pi/2} -4\sin(t) = -4\sin(\pi/2)$$ Since sin(蟺/2) = 1, the limit for the j-component is: $$\lim_{t \rightarrow \pi/2} -4\sin(t) \mathbf{j} = -4\mathbf{j}$$
03

Evaluate the limit for the k-component

The k-component of the given vector function is (2t/蟺). To find its limit as t approaches 蟺/2, substitute 蟺/2 into the function: $$\lim_{t \rightarrow \pi/2} \frac{2t}{\pi} = \frac{2(\pi/2)}{\pi}$$ Simplifying the expression, we get: $$\lim_{t \rightarrow \pi/2} \frac{2t}{\pi} \mathbf{k} =1\mathbf{k}$$
04

Combine the components

Now that we have evaluated the limit for each component, we can combine them to obtain the limit of the entire vector-valued function: $$\lim_{t \rightarrow \pi/2} \left(\cos(2t) \mathbf{i} - 4\sin(t) \mathbf{j} + \frac{2t}{\pi} \mathbf{k}\right) = (-1\mathbf{i} - 4\mathbf{j} + 1\mathbf{k})$$

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Most popular questions from this chapter

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right)$$

Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in \(\mathrm{R}^{3}\) that form the sides of a triangle (see figure). Use the following steps to prove that the medians intersect at a point that divides each median in a 2: 1 ratio. The proof does not use a coordinate system. a. Show that \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\) b. Let \(\mathbf{M}_{1}\) be the median vector from the midpoint of \(\mathbf{u}\) to the opposite vertex. Define \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) similarly. Using the geometry of vector addition show that \(\mathbf{M}_{1}=\mathbf{u} / 2+\mathbf{v} .\) Find analogous expressions for \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) c. Let \(a, b,\) and \(c\) be the vectors from \(O\) to the points one-third of the way along \(\mathbf{M}_{1}, \mathbf{M}_{2},\) and \(\mathbf{M}_{3},\) respectively. Show that \(\mathbf{a}=\mathbf{b}=\mathbf{c}=(\mathbf{u}-\mathbf{w}) / 3\) d. Conclude that the medians intersect at a point that divides each median in a 2: 1 ratio.

Consider the parallelogram with adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\). a. Show that the diagonals of the parallelogram are \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\). b. Prove that the diagonals have the same length if and only if \(\mathbf{u} \cdot \mathbf{v}=0\). c. Show that the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the sides.

The function \(f(x)=\sin n x,\) where \(n\) is a positive real number, has a local maximum at \(x=\pi /(2 n)\) Compute the curvature \(\kappa\) of \(f\) at this point. How does \(\kappa\) vary (if at all) as \(n\) varies?

Note that two lines \(y=m x+b\) and \(y=n x+c\) are orthogonal provided \(m n=-1\) (the slopes are negative reciprocals of each other). Prove that the condition \(m n=-1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v}=0\) where \(\mathbf{u}\) points in the direction of one line and \(\mathbf{v}\) points in the direction of the other line.
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