solve-the-given-problems-by-finding-the-appropriate-derivative-assuming-that-force-is-proportional-3

Question

Solve the given problems by finding the appropriate derivative.
Assuming that force is proportional to acceleration, show that a particle moving along the $x$ -axis, so that its displacement $x=a e^{k t}+b e^{-k t},$ has a force acting on it which is proportional to its displacement.

Stanley Enemuo · Accepted Answer

we need to show that part of what moving along the X axis has a displacement. That is equal to eight times 3 to the K. T. Plus B, times E. To the negative Casey. So let&#x27;s, so the forces that none of my F deceleration by a According to Newton&#x27;s 2nd law efforts, proportion of A. And uh is equal to K. Prone time. Say. So we&#x27;re Kay, prime minister constant of proportionality. So we know that the acceleration is given by the double derivative of displacement. So dx DT That&#x27;s equal to the 18. Uh huh. Two times A. To the K. T minus B. My name&#x27;s beat me. It&#x27;s in a negative Katie. Now, the second derivative, the best case square I&#x27;m E. To the K. T. Must be E to the negative Katie. So put in the value of this place. So X equal to E. To the K. T. Cosby into the -K two.

Robert Zaballa · Answer

we first look at the parallel case in which the forces parallel to the velocity. So in that case, on Lee, the magnitude will change for the velocity. So what we do is we take the time derivative of the moment. Um and so we have the relativistic equation of momentum. We take the time derivative of it, and we&#x27;re actually going to be taking the time derivative of a product of two quantities so we can go ahead and factor out the mass, assuming that it&#x27;s not going to change with time. And we&#x27;re gonna take the derivative of V as well as well as one over the square root of one minus V square over C squared. So we have to use the product rule in calculus for derivatives. And so you see here we&#x27;re gonna have ah one over the square root of one minus V squared over C squared times the time derivative of the of the speed, the velocity or speed plus V times the time derivative of one divided by the square root of one minus V squared over C squared. So we do that and then our next step, our first term is not going to change is just gonna have the same form. In other words, and what I do is I go ahead and rewrite one over the square root of one minus B squared over C squared as one minus V squared overseas where to the minus one half. Because when you take derivatives, you&#x27;re going to use, uh, numerical forms of the powers. So we go ahead and continue and it gets a little bit complicated. We bring down the power negative one half. We bring out the derivative of one minus B squared over C squared. We also have a devi over DT that appears in the second term here. And then we also have one minus V square Oversee square to the minus three half. That also results our first term one over square root of one minus fees wherever c squared times DV over D t remains the same. So as we continue, we can simplify. And we can also get a common denominator between our two terms, namely one minus V square were C square to the three halves. And when we do this, we see that a term an extra term appears in our first term, which is minus V squared over C squared, which cancels with v squared over C squared in the second term. And so our overall quantity reduces Thio The mass times DV over DT times 1/1 minus V squared over C square to the three halves. So this results in M A over one minus V&#x27;s where over C squared to the three halves and in turn we get that. The acceleration for the parallel case is F over M times one minds V Square oversee square to the three has again This is when the forces parallel to the velocity. So now we look at the case when the force and velocity or perpendicular so the velocity is not going to change in magnitude, but it will change in direction. So again, magnitude remains the same. It&#x27;s constant, and also this means that one divided by the square root of one minus V square over C squared is gonna when you take its time, derivative is going to equal zero. So the time derivative of one divided by the square root of one minus V square over C squared equals zero. So the force equation one more time is D over DT, the time derivative of M V over one minus V square over C squared. We can then write this, noting that on Lee the direction changes as mass divided by the square root of one minus V square over C squared times DV over de G. So when we write this, we know that we&#x27;re just We&#x27;re talking about the time rate of change of the direction of the velocity. So the time rate of change of the direction of the velocity. And so we end up with M A divided by the square root of one minus V square or C squared, and so we can write the acceleration as F over M times one minus V squared over C squared. And that means that the magnitude is then a equals f over em times the square root of one minus V squared over C squared

Will Erickson · Answer

either. In this problem, we have to find the derivative of a times E to the B. T. And remember that for all these problems, A and B are just Constance. So treat them just like you wouldn&#x27;t numbers. The only variable is a T. The important thing to see is that we have, um, a times, even the BT. It&#x27;s tempting to use a product rule here, and you could, but we could use a product rule because we do have one function of tea times another. Except this 1st 1 here is not actually a function of tea or there&#x27;s no tea in. There is just a constant. So even though the product will would work, it would end up causing a lot of a lot of terms. There are at least a few terms of zeroes in them and ends up causing a lot of work. So this remembered the constant multiple rule for a derivative, Um, that the derivative if we was multiplying by a that constant just remains in her answer. So we don&#x27;t actually need a product rules because a is not function of tea. Okay, The age of stays on her answer and then the derivative eat of the B T. Well, you just need a chain rule for this. So the derivative of E to any function is itself on eat of that function. But then we need to multiply because of the chain rule. By the derivative of that exponents, it&#x27;s a the derivative of B B time steam. So get a he, the Bt and the derivative of B times T is just because again be is just a constant. So the derivative of B times T is being, and we can write that your final answer is a Times B. I was even the Bt and we&#x27;re done. Hopefully that helped.

Yaw Asomani · Answer

So we&#x27;re continuing the same expedition. Okay, so the A and B are student councils that we&#x27;ve been talking about. Just watched the previous tutorials before, just when you could I understand. So we&#x27;re gonna rewrite this one as a T square, Right? Because this tea here is that the denominator? So he&#x27;s gonna come up. So this is be than t to the negative to write from here. We confined devi the tea, right? So DVD t this two years in a drop to mortify so to a T right, cause this square to your minus one is deal. One city did a power when it&#x27;s still t and then this native to is gonna drop here tomorrow. Surprises would be negative to be right. Do you take away one here? Is that having negative three? Right. And then we&#x27;re gonna wrap it up nicely as negative to be over t Cubic. Right, Because this negative year isn&#x27;t positive here. So still following the same things have been done. We&#x27;ve been doing over and over and over there on the difference here is that now you have generic, uh, constants. Uh, A and B. Right. But it&#x27;s the same thing. Same concept

Will Erickson · Answer

neither for this problem. We have to take this derivative so we will need the product rule. Our first part of the product is T And the second part is e to the minus two tea. So, following the product rule the derivatives, we&#x27;ll be well, it&#x27;ll be the first part which is just tea times the derivative of the second part, plus the derivative of the first part on times just the second part. Thank you. So we have the tea there and the derivative of E to the minus two tea. The derivative Ah e tow Anything is itself. But by the chain rule we need two multiplied by the derivative of negative to t, which is negative too. Okay, Now the derivative of tea is just one. And so we have the answer. Here. We can clean it up just a bit, so we get minus two t e to the minus two tea, plus even minus two tea. Or if you prefer, we can factor out that e t minus two t and we&#x27;re left with minus two t plus one. We&#x27;re done

Eric Mockensturm · Answer

but the acceleration were Particle is defined by the relationship A equals K kinds of quantity one minus e to the minus X where K is a constant, knowing that the velocity of the particle is V equals nine meters per second when X equals minus three meters and that the particle comes to rest at the origin determined a. The value of K and B the velocity of the particle when X equals minus two. So here&#x27;s the acceleration expression that were given. Now we know the problem statement that when X is minus three years, nine meters per second and when x zero via zero, you can then usar integration method here, relating if we have a as A as a function of X and if we have philosophies. So we go from a velocity of 0 to 9 feeding me and then our displacement goes from zero to minus three. And here is our acceleration and we need to integrate this and we can do that and we get that k is 2.52 meters per second squared were then asked um the velocity of the particle when X equals two meters minus two meters So when X is minus two meters, we wind up getting If we do these integral here now we wind up finding that we crime is plus or minus 4.7 meters per second. But assuming it&#x27;s going to leave the plus version since we&#x27;re going from nine down, up to zero.

Problem

Solve the given problems by finding the appropria…

Problem 31 Easy Difficulty

Solve the given problems by finding the appropriate derivative.
Assuming that force is proportional to acceleration, show that a particle moving along the $x$ -axis, so that its displacement $x=a e^{k t}+b e^{-k t},$ has a force acting on it which is proportional to its displacement.

Answer

$$a=k^{2} x$$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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$$a=k^{2} x$$

Basic Technical Mathematics with Calculus

Kristen K.

Samuel H.

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Precalculus Review - Intro

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Kristen K.

Samuel H.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus