AP Calc Unit 4 Study Guide
A particle moves along the y-axis. The graph of the particle's velocity v(t) at time t is shown above for 0<t<4.5. How many times does the particle change direction over the time interval 0<t<4.5 ?
Answer A Correct. The particle's velocity v(t)v(t) changes from negative to positive at t=1t=1 and t=3t=3 and from positive to negative at t=2t=2. Therefore, the particle changes direction from downwards to upwards at t=1t=1 and t=3t=3 and from upwards to downwards at t=2t=2. The particle does not change direction at t=4t=4 because there is no sign change in the velocity.
Charles's law states that if the pressure of a dry gas is held constant, then the volume V of the gas and its temperature T, measured in degrees Kelvin, satisfy the relationship V=cT, where c is a constant. Which of the following best describes the relationship between the rate of change, with respect to time t, of the volume and the rate of change, with respect to time t, of the temperature?
Answer B Correct. Differentiating both sides of V=cTV=cT with respect to tt and using the fact that cc is a constant, gives this equation. In this context, tt and TT represent different quantities, and tt is the independent variable. dTdtdTdt results from application of the chain rule with the derivative of TT.
Two straight roads intersect at right angles. A car is traveling south toward the intersection while a radar-monitored-speed sign is positioned 14 mile east of the intersection. The speed sign provides both the distance from the sign to an approaching car and the rate at which the distance between the sign and the car is changing. At a certain time, the speed sign shows that the approaching car is exactly 12 mile away from the sign and that the rate at which the distance between the car and the sign is decreasing is 60 miles per hour. Which of the following is true about the distance between the car and the intersection at this instant?
Answer B Correct. Let zz denote the distance between the car and the speed sign and yy denote the distance between the approaching car and the intersection. Then z2=y2+(14)2z2=y2+(14)2 and y=14−116−−−−−−√=3√4y=14−116=34 when z=12z=12. Differentiating the first equation with respect to time gives 2zdzdt=2ydydt+02zdzdt=2ydydt+0. Substituting values for yy, zz, and dzdt=−60dzdt=−60, it follows that 2(12)(−60)=2(3√4)dydt+02(12)(−60)=2(34)dydt+0. Therefore, dydt=12(−60)3√4=−1203√dydt=12(−60)34=−1203 miles per hour. Because dydt<0dydt<0, it follows that the distance is decreasing at this instant.
The figure above shows the graph of the twice-differentiable function g and the line tangent to the graph of g at the point (0,3). The value of limx→0g(x)e−x−3x2−2x is
Answer B Correct. Substituting 0 into the fraction yields the indeterminate form 0000 because both the value of the numerator and the value of the denominator are equal to 0. From the graph, g(0)=3g(0)=3. Therefore, L'Hospital's Rule can be used to find the limit. limx→0g(x)e−x−3x2−2x=limx→0g′(x)e−x−g(x)e−x2x−2limx→0g(x)e−x−3x2−2x=limx→0g′(x)e−x−g(x)e−x2x−2 Because g(0)=3g(0)=3 and g′(0)=2g′(0)=2 from the graph, this limit evaluates as 2−3−2=122−3−2=12.
The line tangent to the graph of the twice-differentiable function f at the point x=5 is used to approximate the value of f(5.25). Which of the following statements guarantees that the tangent line approximation at x=5.25 is an overestimate of f(5.25) ?
Answer C Correct. Because the graph is concave down on the interval 5≤x≤5.255≤x≤5.25, the line tangent to the graph at x=5x=5 will lie above the graph of ff over the interval 5≤x≤5.255≤x≤5.25. Therefore the point at x=5.25x=5.25 on the tangent line will be above the point at x=5.25x=5.25 on the graph of ff, and so the tangent line value at x=5.25x=5.25 will be greater than f(5.25)f(5.25). The tangent line approximation is an overestimate when the line is above the curve at the point. The tangent line approximation is an underestimate when the line is below the curve at the point.
Selected values of the twice-differentiable functions f and g and their derivatives are given in the table above. The value of limx→3x3f(x)−54g(x)−1 is
Answer C Correct. Both ff and gg are continuous because they are differentiable. Since limx→3(x3f(x)−54)=27⋅f(3)−54=27⋅2−54=0limx→3(x3f(x)−54)=27⋅f(3)−54=27⋅2−54=0 and limx→3(g(x)−1)=g(3)−1=0limx→3(g(x)−1)=g(3)−1=0, the limit is of the indeterminate form 0000. Therefore, L'Hospital's Rule can be used to find the limit. Both f′f′ and g′g′ are continuous because they are differentiable. limx→3x3f(x)−54g(x)−1=limx→33x2f(x)+f′(x)x3g′(x)=27⋅f(3)+f′(3)⋅27g′(3)=27⋅2+(−4)⋅272=−542=−27
The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan−1(t2), where t is measured in days. What is the instantaneous rate of change of the mass of the colony, in grams per day, at the moment the colony reaches a mass of 6 grams?
Answer C Correct. Find the time when the colony reaches a mass of 6 grams by using the calculator to solve the equation P(t)=2+5tan−1(t2)=6P(t)=2+5tan−1(t2)=6. The solution is t=2.059277t=2.059277. Since the derivative P′P′ can be interpreted as the instantaneous rate of change of PP with respect to time, the calculator is used to numerically calculate the derivative of PP at this value of tt.
Which of the following limits does not yield an indeterminate form?
Answer C Correct. If limx→af(x)=limx→ag(x)=0limx→af(x)=limx→ag(x)=0 or limx→af(x)=limx→ag(x)=∞limx→af(x)=limx→ag(x)=∞, then limx→af(x)g(x)limx→af(x)g(x) is an indeterminate form. Here limx→π(x−π)=0limx→π(x−π)=0 but limx→πcosx=−1limx→πcosx=−1. Therefore, limx→πx−πcosxlimx→πx−πcosx is not an indeterminate form and the value of the limit is 0.
The figure above shows the graph of the differentiable function f for 1≤x≤11 and the secant line through the points (1,f(1)) and (11,f(11)). For how many values of x in the closed interval [1,11] does the instantaneous rate of change of f at x equal the average rate of change of f over that interval?
Answer C Correct. The graph of ff intersects the secant line at x=1x=1, x=ax=a, x=bx=b, and x=11x=11, where 1<a<b<111<a<b<11. In each of the intervals (1,a)(1,a), (a,b)(a,b), and (b,11)(b,11), there will be exactly one value of xx at which the instantaneous rate of change of ff is equal to the average rate of change of ff over the interval [1,11][1,11]. In other words, there will be three values of xx at which the slope of the line tangent to the graph of ff will equal the slope of the secant line.
The derivative of the function A is given by A′(t)=2+9e0.4sint, and A(1.2)=7.5. If the linear approximation to A(t) at t=1.2 is used to estimate A(t), at what value of t does the linear approximation estimate that A(t)=15 ?
Answer C Correct. The linear approximation to A(t)A(t) at t=1.2t=1.2 will be y=A(1.2)+A′(1.2)(t−1.2)y=A(1.2)+A′(1.2)(t−1.2). Since A′(1.2)=2+9e0.4sin1.2=15.06635A′(1.2)=2+9e0.4sin1.2=15.06635, the linear approximation is y=7.5+15.06635(t−1.2)y=7.5+15.06635(t−1.2). The linear approximation will estimate that A(t)=15A(t)=15 when y=7.5+15.06635(t−1.2)=15⇒t=15−7.515.06635+1.2=1.698
A person whose height is M feet is walking away from the base of a streetlight along a straight road, as shown in the figure above. The height of the streetlight is L feet. At time t seconds, the person is x feet from the streetlight, and the length of the person's shadow is s feet. The quantities are related by the equation 1L(x+s)=1Ms, where L and M are constants. Which of the following best describes the relationship between the rate of change of x with respect to time and the rate of change of s with respect to time?
Answer D Correct. Differentiating both sides of the given equation with respect to time yields 1L(dxdt+dsdt)=1Mdsdt1L(dxdt+dsdt)=1Mdsdt. Solving for dxdtdxdt gives dxdt=LMdsdt−dsdtdxdt=LMdsdt−dsdt.
A particle moves on the circle x2+y2=25 in the xy-plane for time t≥0. At the time when the particle is at the point (3,4), dxdt=6. What is the value of dydt at this time?
Answer D Correct. Differentiating both sides of x2+y2=25x2+y2=25 with respect to tt gives 2xdxdt+2ydydt=02xdxdt+2ydydt=0. Substituting in the three known values gives 2(3)(6)+2(4)dydt=02(3)(6)+2(4)dydt=0. Solving for dydtdydt gives the result dydt=−92dydt=−92
A rectangle has width w inches and height h inches, where the width is twice the height. Both w and h are functions of time t, measured in seconds. If A represents the area of the rectangle, which of the following gives the rate of change of A with respect to t ?
Answer D Correct. Given that the width is twice the height, then A=wh=(2h)h=2h2A=wh=(2h)h=2h2. Differentiating with respect to time and using the chain rule gives dAdt=4hdhdtdAdt=4hdhdt. Using the units for hh and dhdtdhdt, the units for dAdtdAdt are in⋅(in/sec)in⋅(in/sec) or in2/secin2/sec.
Oil is spilled onto a kitchen floor. The area covered by the oil at time t is given by the function A, where A(t) is measured in square centimeters and t is measured in seconds. Which of the following gives the rate at which the area covered by the oil is changing at time t=10 ?
Answer D Correct. The derivative of the function AA evaluated at t=10t=10 gives the rate of change of the area covered by the oil at time t=10t=10
A water tank is in the shape of a right circular cone as shown above. The diameter of the cone is 10 feet, and the height is 15 feet. The shape of the water in the tank is conical with radius r feet and height h feet. At noon, water is leaking from the bottom of the tank at a rate of 12 cubic feet per hour, and the volume of water in the tank is 27π cubic feet. At noon, what is the rate at which the height of the water in the tank is changing? (The volume V of a right circular cone with radius r and height h is V=13πr2h.)
Answer D Correct. The radius of the cone is 5 feet. Using similarity, it follows that r=h3r=h3. Let VV denote the volume of the cone of water in the tank. Then V=13πr2h=13π(h3)2h=πh327V=13πr2h=13π(h3)2h=πh327. From this equation, it follows that when V=27πV=27π, h=9h=9. Differentiating with respect to tt, dVdt=πh29dhdtdVdt=πh29dhdt. Substituting values into the differential equation we have −12=92π9dhdt−12=92π9dhdt so dhdt=−43πdhdt=−43π. Therefore, the rate at which the height of the cone of water in the tank is changing is decreasing at 43π43π feet per hour.
The locally linear approximation of the differentiable function f at x=2 is used to approximate the value of f(2.3). The approximation at x=2.3 is an underestimate of the corresponding function value at x=2.3. Which of the following could be the graph of f?
Answer D Correct. The tangent line is the graph of the locally linear approximation of the function ff near the point of tangency at x=2x=2. Since this graph is concave up on the interval (1,3)(1,3), the tangent line would lie under the graph and the tangent line value at x=2.3x=2.3 would be less than f(2.3)f(2.3). Therefore, this could be the graph of ff. The tangent line approximation is an overestimate when the line is above the curve at the point. The tangent line approximation is an underestimate when the line is below the curve at the point.
A particle moves along the x-axis so that at time t≥0 its position is given by x(t)=2t3+3t2−36t+50. What is the total distance traveled by the particle over the time interval 0≤t≤5 ?
Answer D Correct. The velocity of the particle is given by x′(t)=6t2+6t−36=6(t+3)(t−2)x′(t)=6t2+6t−36=6(t+3)(t−2). This is negative for 0≤t<20≤t<2 and positive for 2<t≤52<t≤5. So the particle is moving left over the time interval 0≤t<20≤t<2 and moving right over the time interval 2<t≤52<t≤5. Therefore, the total distance traveled by the particle over the time interval 0≤t≤50≤t≤5 is given by (x(0)−x(2))+(x(5)−x(2))=(50−6)+(195−6)=233(x(0)−x(2))+(x(5)−x(2))=(50−6)+(195−6)=233.
A particle moves along the y-axis so that at time t≥0 its position is given by y(t)=t3−6t2+9t. Over the time interval 0<t<4, for what values of t is the speed of the particle increasing?
Answer D Correct. The velocity of the particle is given by y′(t)=3t2−12t+9=3(t−1)(t−3)y′(t)=3t2−12t+9=3(t−1)(t−3), and the acceleration of the particle is given by y′′(t)=6t−12=6(t−2)y″(t)=6t−12=6(t−2). The speed of the particle is increasing when the velocity and acceleration have the same sign, which occurs for 1<t<21<t<2 and 3<t<43<t<4.