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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 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 distance-time graph is decreasing, or has a negative slope. 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 \(d(t)=3t^3-6t\), then at \(t=2\), the object’s velocity is
Step 1: Velocity is the first derivative of distance. Step 2: Differentiate \(d(t)=3t^3-6t\): \(d'(t)=9t^2-6\). Step 3: To know whether velocity is increasing or decreasing, find acceleration, which is the derivative of velocity. Step 4: Differentiate again: \(d''(t)=18t\). Step 5: Evaluate at \(t=2\): \(d''(2)=18(2)=36\). Step 6: Since \(36>0\), acceleration is positive, so velocity is increasing. Answer: Increasing
3.
Question 3. The power in a circuit is given by \(P=I^2R\), 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 resistance is constant at \(500\), substitute \(R=500\): \(P=500I^2\). Step 3: Differentiate both sides with respect to \(t\): \(\frac{dP}{dt}=500\cdot2I\frac{dI}{dt}\). 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 correct. Answer: True
4.
Question 4. A stone is dropped into a pool and makes a circular ring with area \(A=\pi r^2\). If the radius is increasing at a rate of \(8\text{ cm/sec}\) when the radius is \(30\text{ cm}\), how fast is the area increasing?
Step 1: Start with the area formula \(A=\pi r^2\). Step 2: Differentiate both sides with respect to time \(t\): \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\). Step 3: Identify the given values: \(r=30\) and \(\frac{dr}{dt}=8\). Step 4: Substitute into the formula: \(\frac{dA}{dt}=2\pi(30)(8)\). Step 5: Simplify: \(\frac{dA}{dt}=480\pi\). Step 6: Include the correct units: \(480\pi\text{ cm}^2/\text{sec}\). Answer: \(480\pi\text{ cm}^2/\text{sec}\)
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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 exists \(c\) 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 the interval \((0,\pi)\), the value that satisfies \(\cos c=0\) is \(c=\frac{\pi}{2}\). Answer: \(\frac{\pi}{2}\)
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Question 6. \(\lim_{x\to0}\frac{\cos x-1}{x^2}=\)
Step 1: Substitute \(x=0\) directly: \(\frac{\cos0-1}{0^2}=\frac{1-1}{0}=\frac{0}{0}\), so the limit has indeterminate form \(\frac{0}{0}\). Step 2: Use L'Hôpital's Rule and differentiate the numerator and denominator: \(\lim_{x\to0}\frac{\cos x-1}{x^2}=\lim_{x\to0}\frac{-\sin x}{2x}\). Step 3: Substitute \(x=0\) again: \(\frac{-\sin0}{2(0)}=\frac{0}{0}\), so use L'Hôpital's Rule a second time. Step 4: Differentiate again: \(\lim_{x\to0}\frac{-\sin x}{2x}=\lim_{x\to0}\frac{-\cos x}{2}\). Step 5: Substitute \(x=0\): \(\frac{-\cos0}{2}=\frac{-1}{2}\). Answer: \(-\frac{1}{2}\)
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