{Use of Tech} Spring oscillations A spring hangs from the ce...

Question

{Use of Tech} Spring oscillations A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance x (in inches) of the mass from its equilibrium position after t seconds is given by the function x(t) = 10sin t - 10cos t, where x is positive when the mass is above the equilibrium position. <IMAGE>
b. Find dx/dt and interpret the meaning of this derivative.

b. Find dx/dt and interpret the meaning of this derivative.

Accepted Answer

Hello. In this video, we are going to be approaching the following derivative problem. We are told that a buoy is flowing floating in the sea, and it is bobbing up and down with the waves. The height H in feet of the buoy above the sea level after T seconds is given by the function H of T equal to 3 cosine of T minus 3 sin of T for T greater than or equal to 0. We want to find the derivative of H with respect to T and then provide an interpretation of this derivative. Let's go in and approach this problem by approaching the first part. The first part is asking us to find a derivative of H with respect to the variable T. So, our goal for the first part of the problem is to calculate DHDT. The function given to us is H of T equal to 3 cosine of T. -3 sign of T. So, we are going to take the derivative with respect to the variable T. That is going to require us to know the derivative of cosine and sine. By definition, the derivative of sine is going to be positive cosine. And the derivative of cosine. is going to be negative sign. So by taking the derivative of each of the terms independently, we will have the derivative DHDT equal to 3 multiplied by the derivative of cosine of T, which is negative sine of T. -3 multiplied by the derivative of sin of T, which will be positive cosine of T. And by in the first term, by multiplying the negative coefficient with 3, we can rewrite the derivative to be negative 3, sin of T. -3, cosine of T. And this is going to be the derivative of H with respect to T, which is going to be the first solution that we are looking for. Now what about the second half of this problem? The second half is asking us to provide an interpretation of this derivative. So what it is asking us to do is to describe what DHDT represents in terms of the buoy. Now, DHDT is a derivative with respect to time. So what this represents is the rate of change. Of the height of the buoy. Over any time. Of tea So, if we want to know the rate of change at any time of the buoy, we can plug in any value of T and the derivative will provide us the rate of change at that specific time. With that being said, this is going to be the solution to this problem. So, I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

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