In Exercises $43-46,$ use a CAS to perform the following steps to evaluate the line integrals.

a. Find $d s=|\mathbf{v}(t)| d t$ for the path $\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+$ $k(t) \mathbf{k} .$

b. Express the integrand $f(g(t), h(t), k(t))|\mathbf{v}(t)|$ as a function of the parameter $t .$

c. Evaluate $\int_{C} f d s$ using Equation $(2)$ in the text.

$$f(x, y, z)=\sqrt{1+30 x^{2}+10 y} ; \quad \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+3 t^{2} \mathbf{k} 0 \leq t \leq 2$$

## Discussion

## Video Transcript

Okay, folks. So now, um, we're gonna talk about problem number 39. So we're going to be looking for the Z component of the moment of inertia of a spring of constant density lying along a helix. Um, so for part A, we're just looking for IEDs, e. Basically, um, and that we have a simple expression for that, which is just an integral, um of delta R squared. Minus c squared. Yes. And we're integrating along along the spring. Um, where Delta is the function is the constant density. Now I'm gonna rewrite DS. I'm going to rewrite ds as, um as something else. First of all, our square, my disease, where give you x squared plus y squared. And it ds is just ex dot squared plus y dot squared plus z dot squared multiplied by d. T where I used the notation where dot ex dot gives you dx DT And why doc gives you d y d teensy dog is using these ladies. It's just a convenient location. Um, ex doc. Well, first of all, we have a function for X in terms of tea and function for y in terms of Deanna function, Brazilian tour is empty. Um, this is T and this is sign of T. And this is co sign empty. That's given. And when you when you take you take the time derivative effects and Y and Z, you square them all and you add them up, you're going to get to okay, you're gonna get to and you can check for yourself. So we have we just have a route to hear. This is Route two and a constant Delta's weaken. Pull those two constants out because that's what we do. We can. We can always pull Constance out of the interval. We have route to delta, um, between zero and two. Pi co science query plus sine squared detained. That's very nice, because coup sine squared plus a plus size square always gives you one. So now we have route to Delta D T. Between zero and two pi, and that's very simple to evaluate because it's really just two pi route to delta. So that's for part A. This is the Z component of the moment of inertia, tensor or matrix, Moment of inertia matrix, um, of of this spring lying along the helix. So that's for part a part B seems a little bit longer, but it's really not that much harder. The answer for parties different. It's not the same. The reason is not the same is because because this this thing here is now is now twice as long. So when something is twice as long, that means it's, you know it's heavier. And you know the moment in a moment of inertia, of of an object is a property of the mass distribution of this of this object. And when you have more mass, I think it's pretty intuitive that that this object should have a different, uhm moment of inertia that that's really kind of a hand wavy way of understanding why it should be different. Let's just calculate it out and, um, and let the numbers do the talking. Okay, Anyway, we have I'd see for this for this longer helix or this longer spring. Um, that's really that's really the same thing as this one, except that now we have a different integration limit. And instead of stopping at two pi, we're going to be stopping at four pi. Okay, so basically, we're just gonna copy down this integral here and he raised this to pie and riot four pi because the set up is the same, it's still the same shape. It's still the helix, except now is longer. So now we're gonna have instead of two pi Ruutu Delta. We're gonna have twice that. We're gonna have four pi route to Delta. So, as you can see it is, it is long. I mean, it is different than what we had for part A. And the reason is the reason it's different. It's because now we have a helix that is twice as long is. And thats why is twice as difficult to rotate. Um, something something that's difficult, that is more difficult to rotate, means that it has a bigger moment of inertia. But anyway, that's it for this video. We have our answers for problem number Third night and we're done. Thank you. Goodbye.

## Recommended Questions

A spring of constant density $\delta$ lies along the helix

$$\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 2 \pi$$

a. Find $I_{z}$

b. Suppose that you have another spring of constant density $\delta$ that is twice as long as the spring in part (a) and lies along the helix for $0 \leq t \leq 4 \pi .$ Do you expect $I_{2}$ for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calculating $I_{z}$ for the longer spring.

Predict/Calculate If a mass $m$ is attached to a given spring,

its period of oscillation is $T$ . If two such springs are connected end

to end, and the same mass $m$ is attached, (a) is its period greater

than, less than, or the same as with a single spring? (b) Verify your

answer to part (a) by calculating the new period, $T,$ in terms of

the old period $T .$

Using the mass-spring analogy, predict the behavior as $$t \right arrow+\infty $$ of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem.

(a) $$y^{\prime \prime}+16 y=0 ; \quad y(0)=2, \quad y^{\prime}(0)=0$$

(b)$$y^{\prime \prime}+100 y^{\prime}+y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=0$$

(c)$$$y^{\prime \prime}-6 y^{\prime}+8 y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=0$$

(d)$$y^{\prime \prime}+2 y^{\prime}-3 y=0 ; \quad y(0)=-2, \quad y^{\prime}(0)=0$$

(e)$$y^{\prime \prime}-y^{\prime}-6 y=0 ; \quad y(0)=1, \quad y^{\prime}(0)=1$$

A spring is made of a thin wire twisted into the shape of a circular helix $x=2 \cos t, y=2 \sin t, z=t .$ Find the mass of two turns of the spring if the wire has constant mass density.

Use the formulas $m=\int_{C} \rho d s, \bar{x}=\frac{1}{m} \int_{C} x \rho d s$ $\bar{y}=\frac{1}{m} \int_{c} y \rho d s, I=\int_{C} w^{2} \rho d s.$

Compute the mass $m$ of the helical spring $x=\cos 2 t$ $y=\sin 2 t, z=t, 0 \leq t \leq \pi,$ with density $\rho=z^{2}.$

Variable Spring Constant. As a spring is heated, its spring "constant" decreases. Suppose the spring Is heated so that the spring "constant" at time $t$ is $k(t)=6-t \mathrm{N} / \mathrm{m}$ (see Figure $8.6 ) .$ If the unforced mass-spring system has mass $m=2 \mathrm{kg}$ and a damping constant $b=1 \mathrm{N}$ -sec/m with initial conditions $x(0)=3 \mathrm{m}$and $x^{\prime}(0)=0 \mathrm{m} / \mathrm{sec},$ then the displacement $x(t)$ is governed by the initial value problem $2 x^{\prime \prime}(t)+x^{\prime}(t)+(6-t) x(t)=0$ $x(0)=3, \quad x^{\prime}(0)=0$ Find at least the first four nonzero terms in a power series expansion about $t=0$ for the displacement.

CE Predict/Explain If a mass $m$ is attached to a given spring, its

period of oscillation is $T .$ If two such springs are connected end to

end and the same mass $m$ is attached, (a) is the resulting period of

oscillation greater than, less than, or equal to $T ?$ (b) Choose the

best explanation from among the following:

\begin{equation}

\begin{array}{l}{\text { I. Connecting two springs together makes the spring stiffer, }} \\ {\text { which means that less time is required for an oscillation. }} \\ {\text { II. The period of oscillation does not depend on the length of a }} \\ {\text { spring, only on its force constant and the mass attached to it. }} \\ {\text { The longer spring stretches more easily, and hence takes longer }} \\ {\text { to complete an oscillation. }}\end{array}

\end{equation}

Predict/Calculate A 3.2 -kg mass on a spring oscillates as

shown in the displacement-versus-time graph in Figure $13-41.$

\begin{equation}\begin{array}{l}{\text { (a) Referring to thegraph, atwhat timesbetween } t=0 \text { and } t=6.0 \mathrm{s}} \\ {\text { does the mass experience a force of maximum magnitude? Explain. }}\end{array}\end{equation}

\begin{equation}

\begin{array}{l}{\text { (b) Calculate the magnitude of the maximum force exerted on the }} \\ {\text { mass. (c) At what times shown in the graph does the mass expe- }} \\ {\text { rience zero force? Explain. (d) How much force is exerted on the }} \\ {\text { mass at the time } t=0.50 \text { s? }}\end{array}

\end{equation}

A block of mass $m$ is suspended from two springs having a stiffness of $k_{1}$ and $k_{2},$ arranged a ) parallel to each other, and b) as a series. Determine the equivalent stiffness of a single spring with the same oscillation characteristics and the period of oscillation for each case.

The displacement of a mass suspended on a spring, at time $t,$ is given by $g(t)=$ $-\frac{\sqrt{3}}{2} \sin t+\frac{1}{2} \cos t .$ Find $c$ in the interval $[0,2 \pi)$ such that $g(t)$ can be written in the form $g(t)=\sin (t+c)$.