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Solve the given problems by finding the appropriate derivative.An object on the end of a spring is moving so that its displacement (in $\mathrm{cm}$ ) from the equilibrium position is given by $y=e^{-0.5 t}(0.4 \cos 6 t-0.2 \sin 6 t) .$ Find the expression for thevelocity of the object. What is the velocity when $t=0.26 \mathrm{s} ?$ The motion described by this equation is called damped harmonic motion.

$$v=-e^{-0.5 t}(1.4 \cos 6 t+2.3 \sin 6 t),-2.03 \mathrm{cm} / \mathrm{s}$$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 8

Applications

Derivatives

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Lectures

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In mathematics, precalculu…

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In mathematics, a function…

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Harmonic Motion The displa…

02:45

In Problems $5-8,$ an obje…

00:33

Damped Oscillations. Suppo…

01:40

03:03

03:33

The displacement from equi…

01:34

? Simple Harmonic Motion F…

01:50

Determine the period of th…

02:28

Solve each problem.Mot…

03:20

we need to find the expression for the velocity of the object. This movement of the end of the spring. Well given to displacement and centimeters from the equilibrium position. The equilibrium position is given by the function Y equal to eat the negative zero point five T. Times 0.4 Co sign coastline 16 minus 0.2 time 16. Also we have to find the velocity see when the time 0.26 seconds. So we know the velocity of an object. To find that we have to find the first derivative of the displacement with respect to time. So we know the function. Why? So now we have to find revenue, the white cheese. And we can use integration by parts to find the former by explaining this. So D. Y. T. Is equal to E. to the negative 0.5 T. Times -2.3. Science signed 16 -1.4 co sign 16. And remember that he is 0.26 seconds. Told you plug in 0.26 seconds and sort of derivative function. We have the uh the U. I. D. C. To be equal to negative two points. There are 32 um centimeters per second.

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