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Question 1. For the following graph of velocity (not distance) versus time over the interval \((1.5,5)\), the number of times the acceleration is zero is
Step 1: Acceleration is the derivative of velocity. Step 2: On a velocity-time graph, acceleration equals the slope. Step 3: The slope is zero at horizontal tangents (local maxima/minima). Step 4: The graph has one local minimum and no local maximum on the interval. Step 5: Therefore acceleration is zero once. Answer: 1
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Question 2. If an object travels according to a distance function \(d(t)\), and \(d'(c)<0\), then at \(t=c\) we know
Step 1: The derivative of distance is velocity. Step 2: Since \(d'(c)<0\), the velocity is negative. Step 3: This gives no information about acceleration. Answer: the velocity is negative
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Question 3. If \(z^2=x^2+y^2\), then \(\frac{dz}{dt}=2x+2y\frac{dy}{dt}\).
Step 1: Differentiate implicitly: \(2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\). Step 2: Solve for \(\frac{dz}{dt}\): \(\frac{dz}{dt}=\frac{2x\frac{dx}{dt}+2y\frac{dy}{dt}}{2z}\). Step 3: The given equation is missing terms. Answer: False
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Question 4. A stone is dropped into a pool and makes a circular ring with area \(A=\pi r^2\). If \(\frac{dr}{dt}=8\) cm/s when \(r=30\) cm, how fast is the area increasing?
Step 1: Differentiate \(A=\pi r^2\): \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\). Step 2: Substitute \(r=30\), \(\frac{dr}{dt}=8\). Step 3: \(\frac{dA}{dt}=2\pi(30)(8)=480\pi\). Answer: \(480\pi\)
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Question 5. Could you use L'Hôpital's rule to solve \(\lim_{x\to0^+}\frac{\ln(x+1)}{e^x}\) in its current form?
Step 1: Substitute \(x=0\). Step 2: The limit becomes \(0/1=0\). Step 3: This is not an indeterminate form. Step 4: L'Hôpital's Rule does not apply. Answer: False
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Question 6. The Mean Value Theorem guarantees there is at least one value of \(c\) on \([-3,3]\) such that \(f'(c)=0\) when \(f(x)=x^2+7\).
Step 1: \(f(x)=x^2+7\) is continuous and differentiable. Step 2: Average rate of change on \([-3,3]\) is \((16-16)/6=0\). Step 3: \(f'(x)=2x\). Step 4: Set \(2c=0\), giving \(c=0\in(-3,3)\). Step 5: The Mean Value Theorem guarantees such a value. Answer: True
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