Show that the displacement D(x,t) = ln(ax + bt), where a and b are constants, is a solution to the wave equation. Then find an expression in terms of a and b for the wave speed.

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Hey, everyone in this problem, we're told that Y of XT is equal to E to the exponent negative CX minus DT is the equation of a disturbance in a string where C and D are two positive constant coefficients. They were asked to find the relationship between C and D. So that the equation satisfies the wave equation. And we're given four answer choices here. Option A put C divided by D is equal to one divided by V. Option B is that D divided by C is equal to one divided by B. Option C is that C multiplied by D is equal to one divided by V. And option D is that C multiplied by D is equal to B, all right. So the first thing we wanna do and we want this to satisfy the wave equation. So what is the wave equation? And we'll recall that the wave equation is given by die square Y of XT divided by di X squared is equal to one divided by V squared, multiplied by di squared yxt divided by DY T squared. OK. So we have the second derivative of Y with respect to X is equal to one divided by V squared multiplied by the second derivative of Y with respect to T. So in order to make our equation satisfy this wave equation, we wanna figure out what these second derivatives are, substitute them into this equation and manipulate it in order to get this relationship. OK. So let's start with the first derivatives. OK. So we have the first derivative, we need to find it's gonna be di Y of XT divided by di X is the derivative of Y with respect to X. Now we have an exponential function here. So the derivative is going to just be E to the exponent negative CX minus DT and the function itself, then we need to multiply it K using our chain rule by the derivative of the exponent and the derivative of what's up here. Now, the derivative of the, the exponent with respect to X OK is just going to be negative C. So we have negative C multiplied by E to the exponent negative CX minus DT. And when we're taking the derivative with respect to XT is just a constant. So the derivative of this DT term is just zero. OK. Now we're gonna do the same but for the derivative with respect to T so die Y of XT divided by die whoops diet. What's that? And we're gonna start with the same thing derivative of the exponential is gonna be the exponential itself E to the exponent negative CX minus DT multiplied by the derivative of the exponent. OK. Now, in this case, we're taking the derivative with respect to T. So the X term is gonna be constant, the derivative is gonna be zero. The derivative of this negative DT term is gonna be negative D. And then we multiply that again by a negative. So that's gonna give us positive D. So we get D multiplied by that whole exponent. All right. So we have our first derivatives, we're gonna move to our second derivatives and we're doing the same process. So now we have di squared Y XT divided by di X square. So the second derivative with respect to X and it's the same thing, we have this negative C in front so that constant stays there. And so we have negative C, we're gonna multiply by the exponential. It stays the same. So E to the exponent negative CX minus DT. And then again, we have to take the derivative of what's in the exponent. And once again, that's negative C and so we have negative C multiplied by negative C multiplied by that exponential which gives us C squared E to the exponent negative CX minus DT. Alrighty, one more derivative to calculate, we have the second derivative of Y di squared YXT with respect to tea diet squid. And once again, following the same pattern, we have the constant D up front, we have the exponential term E to the exponent negative CX minus DT. And then our chain rule tells us to take the derivative of that exponent. OK. And we know from the first derivative that the derivative of the exponent is D, we're gonna multiply that in front we have D multiplied by D that gives us D squared, OK. So D squared multiplied by that whole exponential term. And now we have what we need, OK, we have this second derivative with respect X and we have the second derivative with respect to T. And we're gonna substitute those into our wave equation and find the relationship between the two, right. So subbing into the wave equation, what we get is C squared E to the exponent negative CX minus DTK. That second derivative respect to X is equal to one divided by V squared multiplied by D squared E to the negative CX minus DG second derivative of the fourth to now we have that same ex exponential term on both sides. So we can divide by that, we know that that's gonna be nonzero. So there's no issues there. So we can divide by that term and we're left with C squared is equal to one divided by V squared multiplied by D squared. OK. So now we have this relationship between C and D and we just need to manipulate it to look like one of our solutions. So if we look at the answer choices, we can see that we want the C and D on one side and the V term on the other side. OK. So let's look at our equation try to rearrange so that we can get that to happen, we can divide both sides by D squared. That's gonna give us C squared, divided by D squared is equal to one divided by V squared. If we take the square root of both sides, we get C divided by D is equal to one divided by V. And that's exactly one of those answer choices that we have. OK. So this is the relationship we were looking for. So that ratio of C to D is gonna be the same as the ratio of one to B. All right. So if we go back to our answer choices, we can see that this corresponds with answer choice A, OK? C divided by D is equal to one divided by V. Thanks everyone for watching. I hope this video helped see you in the next one.

Show that the displacement D(x,t) = ln(ax + bt), where a and b are constants, is a solution to the wave equation. Then find an expression in terms of a and b for the wave speed.

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Key Concepts

Wave Equation

Displacement Function

Wave Speed