AP Calculus Test: Limits And Continuity!

1. A triangle has sides a = 9.0 cm, b = 8.0 cm and c = 6.0 cm. Angle A is equal to:

82.42°

56.49°

78.58°

79.87°

The answer is 78.58°. This can be determined by using the Law of Cosines, which states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. By plugging in the given side lengths and solving for angle A, we find that angle A is approximately 78.58°.

Explanation

The answer is 78.58°. This can be determined by using the Law of Cosines, which states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. By plugging in the given side lengths and solving for angle A, we find that angle A is approximately 78.58°.

2. An arc of a circle of length 5.0 cm subtends an angle of 2 radians. The circumference of the circle is:

2.5 cm

10.0 cm

5.0 cm

15.7 cm

The circumference of a circle is calculated by multiplying the radius by 2π. In this case, we are given the length of an arc and the angle it subtends. The formula for finding the length of an arc is L = rθ, where L is the length of the arc, r is the radius, and θ is the angle in radians. We are given L = 5.0 cm and θ = 2 radians. Rearranging the formula, we get r = L/θ. Substituting the given values, we get r = 5.0 cm/2 = 2.5 cm. The circumference of the circle is then calculated as 2πr = 2π(2.5 cm) = 15.7 cm.

Explanation

The circumference of a circle is calculated by multiplying the radius by 2π. In this case, we are given the length of an arc and the angle it subtends. The formula for finding the length of an arc is L = rθ, where L is the length of the arc, r is the radius, and θ is the angle in radians. We are given L = 5.0 cm and θ = 2 radians. Rearranging the formula, we get r = L/θ. Substituting the given values, we get r = 5.0 cm/2 = 2.5 cm. The circumference of the circle is then calculated as 2πr = 2π(2.5 cm) = 15.7 cm.

3. The mean value of a Sine wave over half a cycle is:

0.318 × maximum value

0.707 × maximum value

The peak value

0.637 × maximum value

The mean value of a sine wave over half a cycle is 0.637 times the maximum value. This can be derived from the mathematical properties of a sine wave. The mean value is calculated by finding the average value of the wave over a given interval. In the case of half a cycle, the wave starts at 0, reaches its maximum value, and then returns to 0. The average value of this interval is 0.637 times the maximum value.

Explanation

The mean value of a sine wave over half a cycle is 0.637 times the maximum value. This can be derived from the mathematical properties of a sine wave. The mean value is calculated by finding the average value of the wave over a given interval. In the case of half a cycle, the wave starts at 0, reaches its maximum value, and then returns to 0. The average value of this interval is 0.637 times the maximum value.

4. Four engineers can complete a task in 5 hours. Assuming the rate of work remains constant, six engineers will complete the task in:

126h

4h 48min

3h 20min

7h 30min

If four engineers can complete a task in 5 hours, it means that the total work required for the task is divided equally among the four engineers and each engineer takes 5 hours to complete their portion of the work. Therefore, the total work required for the task can be calculated as 4 engineers x 5 hours = 20 engineer-hours.

If six engineers are working on the same task, the total work required remains the same at 20 engineer-hours. Since the rate of work remains constant, each engineer will still take the same amount of time to complete their portion of the work. Therefore, the time taken by each engineer can be calculated as 20 engineer-hours divided by 6 engineers, which equals approximately 3 hours and 20 minutes.

Explanation

If four engineers can complete a task in 5 hours, it means that the total work required for the task is divided equally among the four engineers and each engineer takes 5 hours to complete their portion of the work. Therefore, the total work required for the task can be calculated as 4 engineers x 5 hours = 20 engineer-hours.

If six engineers are working on the same task, the total work required remains the same at 20 engineer-hours. Since the rate of work remains constant, each engineer will still take the same amount of time to complete their portion of the work. Therefore, the time taken by each engineer can be calculated as 20 engineer-hours divided by 6 engineers, which equals approximately 3 hours and 20 minutes.

5. 11 mm expressed as a percentage of 41 mm is:

2.68, correct to 3 significant figures

2.6, correct to 2 significant figures

26.83, correct to 2 decimal places

0.2682, correct to 4 decimal places

To express 11 mm as a percentage of 41 mm, we can use the formula: (11/41) * 100. Evaluating this expression gives us 26.82926829. Rounding this to 2 decimal places gives us 26.83, which is the correct answer.

Explanation

To express 11 mm as a percentage of 41 mm, we can use the formula: (11/41) * 100. Evaluating this expression gives us 26.82926829. Rounding this to 2 decimal places gives us 26.83, which is the correct answer.

6. The relationship between the temperature in degrees Fahrenheit (F) and the temperature in degrees Celsius (C) is given by: F = 9/5C + 32. 135◦F is equivalent to:

43◦C

57.2◦C

185.4◦C

184◦C

To convert Fahrenheit to Celsius, we subtract 32 from the Fahrenheit temperature and then multiply the result by 5/9. In this case, to find the equivalent Celsius temperature for 135°F, we subtract 32 from 135, which gives us 103. Then we multiply 103 by 5/9, which equals 57.2. Therefore, 135°F is equivalent to 57.2°C.

Explanation

To convert Fahrenheit to Celsius, we subtract 32 from the Fahrenheit temperature and then multiply the result by 5/9. In this case, to find the equivalent Celsius temperature for 135°F, we subtract 32 from 135, which gives us 103. Then we multiply 103 by 5/9, which equals 57.2. Therefore, 135°F is equivalent to 57.2°C.

7. The vertical displacement, s, of a prototype model in a tank is given by s = 40 sin 0.1t mm, where t is the time in seconds. The vertical velocity of the model, in mm/s, is:

− cos 0.1t

400 cos 0.1t

−400 cos 0.1t

4 cos 0.1t

The vertical velocity of an object is the derivative of its vertical displacement with respect to time. Taking the derivative of the given displacement equation, we get ds/dt = 40 * 0.1 * cos(0.1t) mm/s. Simplifying this expression, we have ds/dt = 4 cos(0.1t) mm/s. Therefore, the correct answer is 4 cos(0.1t).

Explanation

The vertical velocity of an object is the derivative of its vertical displacement with respect to time. Taking the derivative of the given displacement equation, we get ds/dt = 40 * 0.1 * cos(0.1t) mm/s. Simplifying this expression, we have ds/dt = 4 cos(0.1t) mm/s. Therefore, the correct answer is 4 cos(0.1t).

8. A vehicle has a velocity v = (2 + 3t) m/s after t seconds. The distance travelled is equal to the area under the v/t graph. In the first 3 seconds the vehicle has travelled:

11m

33m

13.5m

19.5m

The distance travelled is equal to the area under the v/t graph. In this case, the velocity is given by v = (2 + 3t) m/s. To find the distance travelled in the first 3 seconds, we need to find the area under the graph from t=0 to t=3. This can be done by finding the integral of the velocity function with respect to time over the interval [0, 3].

Integrating v = (2 + 3t) with respect to t gives us the distance function d = 2t + (3/2)t^2.

Evaluating this function at t=3 gives us d = 2(3) + (3/2)(3^2) = 6 + (3/2)(9) = 6 + 13.5 = 19.5m. Therefore, the vehicle has travelled 19.5m in the first 3 seconds.

Explanation

The distance travelled is equal to the area under the v/t graph. In this case, the velocity is given by v = (2 + 3t) m/s. To find the distance travelled in the first 3 seconds, we need to find the area under the graph from t=0 to t=3. This can be done by finding the integral of the velocity function with respect to time over the interval [0, 3].

Integrating v = (2 + 3t) with respect to t gives us the distance function d = 2t + (3/2)t^2.

Evaluating this function at t=3 gives us d = 2(3) + (3/2)(3^2) = 6 + (3/2)(9) = 6 + 13.5 = 19.5m. Therefore, the vehicle has travelled 19.5m in the first 3 seconds.

9. The velocity of a car (in m/s) is related to time t seconds by the equation v = 4.5 + 18t − 4.5t 2. The maximum speed of the car, in km/h, is:

81

6.25

22.5

77

The equation v = 4.5 + 18t - 4.5t^2 represents the velocity of the car as a function of time. To find the maximum speed of the car, we need to find the maximum value of this function. This can be done by taking the derivative of the function and setting it equal to zero. By solving this equation, we can find the value of t at which the car reaches its maximum speed. Substituting this value of t back into the original equation, we can find the maximum speed of the car. In this case, the maximum speed is 22.5 km/h.

Explanation

The equation v = 4.5 + 18t - 4.5t^2 represents the velocity of the car as a function of time. To find the maximum speed of the car, we need to find the maximum value of this function. This can be done by taking the derivative of the function and setting it equal to zero. By solving this equation, we can find the value of t at which the car reaches its maximum speed. Substituting this value of t back into the original equation, we can find the maximum speed of the car. In this case, the maximum speed is 22.5 km/h.

10. An alternating current is given by i = 4 sin 150t amperes, where t is the time in seconds. The rate of change of current at t = 0.025s is:

3.99A/s

−492.3A/s

−3.28A/s

598.7A/s

The given equation represents an alternating current with the form i = 4 sin 150t. To find the rate of change of current at t = 0.025s, we need to take the derivative of the equation with respect to time. The derivative of sin is cos, and the derivative of 150t is 150. Therefore, the rate of change of current is given by the derivative of the equation, which is di/dt = 150 * 4 * cos 150t. Evaluating this at t = 0.025s gives di/dt = 150 * 4 * cos 150 * 0.025 = 150 * 4 * cos 3.75. Evaluating this expression gives di/dt ≈ -3.28A/s, which matches the given answer.

Explanation

The given equation represents an alternating current with the form i = 4 sin 150t. To find the rate of change of current at t = 0.025s, we need to take the derivative of the equation with respect to time. The derivative of sin is cos, and the derivative of 150t is 150. Therefore, the rate of change of current is given by the derivative of the equation, which is di/dt = 150 * 4 * cos 150t. Evaluating this at t = 0.025s gives di/dt = 150 * 4 * cos 150 * 0.025 = 150 * 4 * cos 3.75. Evaluating this expression gives di/dt ≈ -3.28A/s, which matches the given answer.