1.
Question 1. For the following graph of distance \(y\)-axis versus time \(x\)-axis, the particle's velocity is negative during the interval \((-\infty,1)\).
Step 1: Velocity is the first derivative of distance with respect to time. Step 2: On a distance-time graph, velocity is represented by the slope of the graph. Step 3: A negative velocity means the graph has a negative slope, or is decreasing. Step 4: Looking at the given graph, the graph is decreasing on the interval \((-\infty,1)\). Step 5: Therefore, the velocity is negative during that interval. Answer: True
2.
Question 2. If the distance of an object is given by the function \(d(t)=3t^2-5t\), then the acceleration of the object at \(t=3\) is
Step 1: Velocity is the first derivative of distance. Step 2: Differentiate \(d(t)=3t^2-5t\): \(d'(t)=6t-5\). Step 3: Acceleration is the derivative of velocity, or the second derivative of distance. Step 4: Differentiate again: \(d''(t)=6\). Step 5: Since \(d''(t)=6\) for all \(t\), the acceleration at \(t=3\) is also \(6\). Answer: 6
3.
Question 3. The power in a circuit is given by the formula \(P=I^2R\), where \(I\) is the current and \(R\) is the resistance. If the resistance is always constant at \(500\), the current is \(1\), and \(\frac{dI}{dt}=-0.5\), then \(\frac{dP}{dt}=-500\).
Step 1: Start with the formula \(P=I^2R\). Step 2: Since the resistance is constant at \(500\), substitute \(R=500\): \(P=500I^2\). Step 3: Differentiate both sides with respect to time: \(\frac{dP}{dt}=500\cdot2I\frac{dI}{dt}\). Step 4: Simplify: \(\frac{dP}{dt}=1000I\frac{dI}{dt}\). Step 5: Substitute \(I=1\) and \(\frac{dI}{dt}=-0.5\): \(\frac{dP}{dt}=1000(1)(-0.5)=-500\). Step 6: The statement is true. Answer: True
4.
Question 4. A \(6\text{ ft}\) ladder, resting against a wall, begins to slip down the wall. When the angle of the ladder is \(45^\circ\), the bottom of the ladder is moving away from the wall at \(1.6\text{ ft/s}\). At that moment, how fast is the top of ladder moving down the wall?
Step 1: Let \(x\) be the distance from the wall to the bottom of the ladder, and let \(y\) be the height of the top of the ladder on the wall. Step 2: Since the ladder is \(6\text{ ft}\) long, \(x^2+y^2=6^2=36\). Step 3: Differentiate both sides with respect to time: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). Step 4: At \(45^\circ\), the triangle has equal legs, so \(x=y\). Step 5: The bottom is moving away from the wall, so \(\frac{dx}{dt}=1.6\). Step 6: Substitute \(x=y\): \(2x(1.6)+2x\frac{dy}{dt}=0\). Step 7: Solve: \(3.2x+2x\frac{dy}{dt}=0\), so \(\frac{dy}{dt}=-1.6\). Step 8: The negative sign means the top is moving downward. The speed is \(1.6\text{ ft/s}\). Step 9: Since \(1.6\) is between \(1.3\) and \(2.0\), the correct answer is between \(1.3\) and \(2.0\text{ ft/s}\). Answer: between \(1.3\) and \(2.0\text{ ft/s}\)
5.
Question 5. Find the value of \(c\) for which the function \(f(x)=\sin x\) on the interval \([0,\pi]\) satisfies the Mean Value Theorem.
Step 1: The Mean Value Theorem says there is a value \(c\) in \((a,b)\) such that \(f'(c)=\frac{f(b)-f(a)}{b-a}\). Step 2: Here \(a=0\), \(b=\pi\), and \(f(x)=\sin x\). Step 3: Compute the average rate of change: \(\frac{f(\pi)-f(0)}{\pi-0}=\frac{\sin\pi-\sin0}{\pi}=\frac{0-0}{\pi}=0\). Step 4: Find the derivative: \(f'(x)=\cos x\). Step 5: Set \(f'(c)=0\): \(\cos c=0\). Step 6: On \((0,\pi)\), the value satisfying \(\cos c=0\) is \(c=\frac{\pi}{2}\). Answer: \(\frac{\pi}{2}\)
6.
Question 6. \(\lim_{x\to0}\ln(x+1)\csc x=\)
Step 1: Rewrite \(\csc x\) as \(\frac{1}{\sin x}\). Then \(\ln(x+1)\csc x=\frac{\ln(x+1)}{\sin x}\). Step 2: Substitute \(x=0\): \(\frac{\ln(1)}{\sin0}=\frac{0}{0}\), which is an indeterminate form. Step 3: Use L'Hôpital's Rule. Differentiate the numerator and denominator: \(\frac{d}{dx}\ln(x+1)=\frac{1}{x+1}\) and \(\frac{d}{dx}\sin x=\cos x\). Step 4: The limit becomes \(\lim_{x\to0}\frac{1}{(x+1)\cos x}\). Step 5: Substitute \(x=0\): \(\frac{1}{(0+1)\cos0}=\frac{1}{1\cdot1}=1\). Answer: 1
1 out of 1
