Question

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Answer 1

The function given is:

`h(t)=e^(-0.3t)\sin(5t)`

To find the first derivative, we need to use several rules of differentiation.

One of them is the chain rule.

Given two functions f(x) and g(x), the chain rule is:

`(d)/(dx)f(g(x))=f^(\prime)(g(x))g^(\prime)(x)`

Next, the derivative of the exponential:

`(d)/(dx)(e^x)=e^x`

The derivative of the product:

`(d)/(dx)(f\cdot g)(x)=f^(\prime)(x)g(x)+f(x)g^(\prime)(x)`

And the derivative of sine and cosine:

`\begin{gathered} (d)/(dx)(\sin(x))=\cos(x) \\ . \\ (d)/(dx)(\cos(x))=-\sin(x) \end{gathered}`

Now, we can break the function into pieces. First we have:

`e^(-0.3t)`

The derivative is, by the chin rule:

:

`(d(e^(-0.3t)))/(dt)=e^(-0.3t)\cdot(d(-0.3t))/(dt)=e^(-0.3t)\cdot(-0.3)=-0.3e^(-0.3t)`

Next:

`(d)/(dt)\sin(5t)=\cos(5t)\cdot(d(5t))/(dt)=5\cos(5t)`

Then, we can write:

`(dh)/(dt)=(d)/(dt)(e^(-0.3t))\cdot\sin(5t)+e^(-0.3t)\cdot(d)/(dx)\sin(5t)=-0.3e^(-0.3t)\sin(5t)+e^(-0.3t)\cdot5\cos(5t)`

We can simplify:

`(dh)/(dt)=5e^(-0.3t)\cos(5t)-0.3e^(-0.3t)\sin(5t)`

Now, for the next derivative, we can separate the terms:

`5e^(-0.3t)\cos(5t)`

And differentiate:

`(d)/(dt)(5e^(-0.3t)\cos(5t))=5((d(e^(-0.3t)))/(dt)\cdot\cos(5t)+e^(-0.3t)\cdot(d)/(dt)(\cos(5t)))=5((-0.3e^(-0.3t))\cdot\cos(5t)+e^(-0.3t)(-5\sin(5t)))`

We can simplify:

`-1.5e^(-0.3t)\cos(5t)-25e^(-0.3t)\sin(5t)`

The other term of dh/dt is:

`(d)/(dx)(-0.3e^(-0.3t)\sin(5t))=-0.3((d)/(dx)(e^(-0.3t))\sin(5t)+e^(-0.3t)\cdot(d)/(dx)(\sin(5t))=-0.3(-0.3e^(-0.3t)\sin(5t)+e^(-0.3t)5\cos(5t))`

Simplify.

`0.09e^(-0.3t)\sin(5t)-1.5e^(-0.3t)\cos(5t)`

Thus:

`(d^2h)/(dt^2)=-1.5e^(-0.3t)\cos(5t)-25e^(-0.3t)\sin(5t)+0.09e^(-0.3t)\sin(5t)-1.5e^(-0.3t)\cos(5t)`

We can simplify:

`e^(-0.3t)(-1.5\cos(5t)-1.5\cos(5t)-25\sin(5t)+0.09\sin(5t))=e^(-0.3t)(-3\cos(5t)-24.91\sin(5t))=-e^(-0.3t)(3\cos(5t)+24.91\sin(5t))`

The answer is:

`(d^2h)/(dt^2)=-e^(-0.3t)(3\cos(5t)+24.91\sin(5t))`

Finally, the second derivative of a position function represents the acceleration, thus:

The correct measurement unit for the second derivative is: cm/sec^2

Answers:

`(dh)/(dt)=e^(-0.3t)(5\cos(5t)-0.3\sin(5t))`
`(d^(2)h)/(dt^(2))=-e^(-0.3t)(3\cos(5t)+24.91\sin(5t))`

The correct measurement unit for the second derivative is: cm/sec^2