The function given is:
[tex]h(t)=e^{-0.3t}\sin(5t)[/tex]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:
[tex]\frac{d}{dx}f(g(x))=f^{\prime}(g(x))g^{\prime}(x)[/tex]Next, the derivative of the exponential:
[tex]\frac{d}{dx}(e^x)=e^x[/tex]The derivative of the product:
[tex]\frac{d}{dx}(f\cdot g)(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)[/tex]And the derivative of sine and cosine:
[tex]\begin{gathered} \frac{d}{dx}(\sin(x))=\cos(x) \\ . \\ \frac{d}{dx}(\cos(x))=-\sin(x) \end{gathered}[/tex]Now, we can break the function into pieces. First we have:
[tex]e^{-0.3t}[/tex]The derivative is, by the chin rule:
:
[tex]\frac{d(e^{-0.3t})}{dt}=e^{-0.3t}\cdot\frac{d(-0.3t)}{dt}=e^{-0.3t}\cdot(-0.3)=-0.3e^{-0.3t}[/tex]Next:
[tex]\frac{d}{dt}\sin(5t)=\cos(5t)\cdot\frac{d(5t)}{dt}=5\cos(5t)[/tex]Then, we can write:
[tex]\frac{dh}{dt}=\frac{d}{dt}(e^{-0.3t})\cdot\sin(5t)+e^{-0.3t}\cdot\frac{d}{dx}\sin(5t)=-0.3e^{-0.3t}\sin(5t)+e^{-0.3t}\cdot5\cos(5t)[/tex]We can simplify:
[tex]\frac{dh}{dt}=5e^{-0.3t}\cos(5t)-0.3e^{-0.3t}\sin(5t)[/tex]Now, for the next derivative, we can separate the terms:
[tex]5e^{-0.3t}\cos(5t)[/tex]And differentiate:
[tex]\frac{d}{dt}(5e^{-0.3t}\cos(5t))=5(\frac{d(e^{-0.3t})}{dt}\cdot\cos(5t)+e^{-0.3t}\cdot\frac{d}{dt}(\cos(5t)))=5((-0.3e^{-0.3t})\cdot\cos(5t)+e^{-0.3t}(-5\sin(5t)))[/tex]We can simplify:
[tex]-1.5e^{-0.3t}\cos(5t)-25e^{-0.3t}\sin(5t)[/tex]The other term of dh/dt is:
[tex]\frac{d}{dx}(-0.3e^{-0.3t}\sin(5t))=-0.3(\frac{d}{dx}(e^{-0.3t})\sin(5t)+e^{-0.3t}\cdot\frac{d}{dx}(\sin(5t))=-0.3(-0.3e^{-0.3t}\sin(5t)+e^{-0.3t}5\cos(5t))[/tex]Simplify.
[tex]0.09e^{-0.3t}\sin(5t)-1.5e^{-0.3t}\cos(5t)[/tex]Thus:
[tex]\frac{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)[/tex]We can simplify:
[tex]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))[/tex]The answer is:
[tex]\frac{d^2h}{dt^2}=-e^{-0.3t}(3\cos(5t)+24.91\sin(5t))[/tex]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:
[tex]\frac{dh}{dt}=e^{-0.3t}(5\cos(5t)-0.3\sin(5t))[/tex][tex]\frac{d^{2}h}{dt^{2}}=-e^{-0.3t}(3\cos(5t)+24.91\sin(5t))[/tex]
The correct measurement unit for the second derivative is: cm/sec^2
