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Question 1. 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 a distance or position function represents velocity. Step 2: Since the distance function is \(d(t)\), its derivative is \(d'(t)\), which gives the velocity at time \(t\). Step 3: We are told that \(d'(c)<0\). Step 4: This means the velocity at \(t=c\) is negative. Step 5: This does not tell us whether acceleration is positive or negative, because acceleration would require the second derivative \(d''(c)\). Answer: the velocity is negative
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Question 2. 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 with respect to time. Step 2: On a velocity-time graph, acceleration is represented by the slope of the graph. Step 3: Acceleration is zero when the slope of the velocity-time graph is zero. Step 4: A horizontal tangent occurs at local maxima or local minima of the velocity graph. Step 5: Looking at the given graph over \((1.5,5)\), there is one local minimum and no local maximum. Step 6: Therefore, the acceleration is zero exactly one time. Answer: 1
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Question 3. If \(z^2=x^2+y^2\), then \(\frac{dz}{dt}=2x+2y\frac{dy}{dt}\).
Step 1: Start with the equation \(z^2=x^2+y^2\). Step 2: Differentiate both sides with respect to \(t\). Step 3: The derivative of \(z^2\) is \(2z\frac{dz}{dt}\). Step 4: The derivative of \(x^2+y^2\) is \(2x\frac{dx}{dt}+2y\frac{dy}{dt}\). Step 5: Therefore, \(2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\). Step 6: Solving for \(\frac{dz}{dt}\), we get \(\frac{dz}{dt}=\frac{2x\frac{dx}{dt}+2y\frac{dy}{dt}}{2z}\). Step 7: The statement \(\frac{dz}{dt}=2x+2y\frac{dy}{dt}\) is missing factors and division by \(2z\), so it is incorrect. Answer: False
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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, the relationship is \(x^2+y^2=6^2=36\). Step 3: Differentiate both sides with respect to time: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). Step 4: At \(45^\circ\), the right triangle has equal legs, so \(x=y\). Step 5: The bottom is moving away from the wall at \(\frac{dx}{dt}=1.6\). Step 6: Substitute \(x=y\) and \(\frac{dx}{dt}=1.6\) into the differentiated equation: \(2x(1.6)+2x\frac{dy}{dt}=0\). Step 7: Solve: \(3.2x+2x\frac{dy}{dt}=0\), so \(2x\frac{dy}{dt}=-3.2x\). Step 8: Divide by \(2x\): \(\frac{dy}{dt}=-1.6\). Step 9: The negative sign means the top is moving downward, and the speed is \(1.6\text{ ft/s}\). Step 10: Since \(1.6\) is between \(1.3\) and \(2.0\), that is the correct interval. Answer: between \(1.3\) and \(2.0\text{ ft/s}\)
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Question 5. The Mean Value Theorem can be used on the function \(f(x)=\frac{1}{x^4}\) on the interval \([-2,2]\).
Step 1: The Mean Value Theorem requires the function to be continuous on the closed interval \([a,b]\). Step 2: It also requires the function to be differentiable on the open interval \((a,b)\). Step 3: The function is \(f(x)=\frac{1}{x^4}\). Step 4: This function is undefined at \(x=0\), because the denominator becomes \(0\). Step 5: Since \(0\) is inside the interval \([-2,2]\), the function is not continuous on the entire interval. Step 6: Therefore, the Mean Value Theorem cannot be used on this interval. Answer: False
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Question 6. Could you use L’Hopital’s rule to solve the following question in its current form: \(\lim_{x\to0^+}\frac{\ln(x+1)}{e^x}\)?
Step 1: L’Hopital’s Rule can only be used directly when the limit gives an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Step 2: Substitute \(x=0\) into the expression. Step 3: The numerator becomes \(\ln(0+1)=\ln 1=0\). Step 4: The denominator becomes \(e^0=1\). Step 5: The limit gives \(\frac{0}{1}=0\), which is not an indeterminate form. Step 6: Since the limit is already determined by direct substitution, L’Hopital’s Rule cannot be used in its current form. Answer: False
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