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| − | <span class="exam">The displacement from equilibrium of an object in harmonic motion on the end of a spring is: | + | <span class="exam"> Find the derivatives of the following functions. Do not simplify. |
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| − | ::<span class="exam"><math>y=\frac{1}{3}\cos(12t)-\frac{1}{4}\sin(12t)</math>
| + | <span class="exam">(a) <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math> |
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| − | <span class="exam">where <math style="vertical-align: -4px">y</math> is measured in feet and <math style="vertical-align: 0px">t</math> is the time in seconds. | + | <span class="exam">(b) <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math> |
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| − | <span class="exam">Determine the position and velocity of the object when <math style="vertical-align: -14px">t=\frac{\pi}{8}.</math> | + | <span class="exam">(c) <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math> |
| | + | <hr> |
| | + | [[009A Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Foundations:
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| − | |What is the relationship between the position <math style="vertical-align: -5px">s(t)</math> and the velocity <math style="vertical-align: -5px">v(t)</math> of an object?
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| − | | <math>v(t)=s'(t)</math>
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| − | |}
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |To find the position of the object at <math>t=\frac{\pi}{8},</math>
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| − | |-
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| − | |we need to plug <math>t=\frac{\pi}{8}</math> into the equation <math style="vertical-align: -5px">y.</math>
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| − | |Thus, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{y\bigg(\frac{\pi}{8}\bigg)} & = & \displaystyle{\frac{1}{3}\cos\bigg(\frac{12\pi}{8}\bigg)-\frac{1}{4}\sin\bigg(\frac{12\pi}{8}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{3}\cos\bigg(\frac{3\pi}{2}\bigg)-\frac{1}{4}\sin\bigg(\frac{3\pi}{2}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{0-\frac{1}{4}(-1)}\\
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| − | &&\\
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| − | &= & \displaystyle{\frac{1}{4} \text{ ft}.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, to find the velocity function, we need to take the derivative of the position function.
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| − | |Thus, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{v(t)} & = & \displaystyle{y'}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-1}{3}\sin(12t)(12)-\frac{1}{4}\cos(12t)(12)}\\
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| − | &&\\
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| − | & = & \displaystyle{-4\sin(12t)-3\cos(12t).}
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| − | \end{array}</math>
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| − | |Therefore, the velocity of the object at time <math>t=\frac{\pi}{8}</math> is
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{v\bigg(\frac{\pi}{8}\bigg)} & = & \displaystyle{-4\sin\bigg(\frac{3\pi}{2}\bigg)-3\cos\bigg(\frac{3\pi}{2}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{-4(-1)+0}\\
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| − | &&\\
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| − | & = & \displaystyle{4 \text{ ft/sec}.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | position is <math>\frac{1}{4} \text{ ft}.</math>
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| − | |-
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| − | | velocity is <math>4 \text{ ft/sec}.</math>
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| − | |}
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| | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
