3.The position (feet traveled) of a car is given by the equation s(t)= 4t2 + 4t. Find the time when the car is going the same speed as its average speed over the interval 0 to 10 seconds.
A)t=0
B)t=2.5
C)t=5
D)t=10
E)Never
4. Consider the curve given by x2 + sin(xy) + 3y2 = C, where C is a constant. The point (1, 1) lies on this curve. Use the tangent line approximation to approximate the y-coordinate when x = 1.01.
A)0.996
B)1
C)1.004
D) Cannot be determined
E)1.338
5.The volume of an open-topped box with a square base is 245 cubic centimeters. Find the height, in centimeters, of the box that uses the least amount of material.
A)7.883 Cm
B) 6 cm
C)3.942 cm
D) 3 cm
E) 2 cm
EITHER:
s' = ds/dt
s' = 8t + 4 = speed in ft/sec
At t = 0, speed = 4 ft/sec
At time t =10, speed = 84 ft/sec
Avg Speed = 0.5(84 + 4) = 44 ft/sec
OR:
At time t = 0, s = 0 ft
At time t = 10, s = 440 ft
Avg. Speed = 440 ft/10 sec = 44 ft/sec
Prob. 5:
Let base measure b×b
Let height = h
Vol = h × b^2 = 245 cm^3
Material = 2×b^2 + 4×bh
b = sqrt(245/h)
Material = 2×(245/h)+4×sqrt(245h)
SOLVE: d(Material)/dh = 0 to find h.
Differentiate the given equation to obtain:
2xdx+cos(xy)(xdy + ydx)+6ydy = 0
2xdx+cos(xy)ydx = -cos(xy)xdy-6ydy
dy = (2xdx+cos(xy)ydx)/(-cos(xy)x-6y)
At (x,y) = (1,1), and dx =0.01 we find:
dy = [2×(0.01)+0.01×cos(1)]/[-cos(1) - 6]
cos(1) = 0.54 so that
dy = .025/-5.46 = -0.005 so that the y coordinate is 1-.005 = .995 when x =.01.
Check this!
3. To find the time when the car is going the same speed as its average speed over the interval 0 to 10 seconds, we need to find the derivative of the position function and then calculate the average speed of the car over the interval.
Let's start by finding the derivative of the position function s(t) = 4t^2 + 4t. The derivative will give us the velocity function, which represents the car's speed at any given time.
ds/dt = 8t + 4
Next, we need to calculate the average speed of the car over the interval 0 to 10 seconds. The average speed is given by the total distance traveled divided by the total time taken.
Average speed = (s(10) - s(0)) / (10 - 0)
Substituting the values into the equation:
Average speed = (4(10)^2 + 4(10) - 4(0)^2 - 4(0)) / 10
Simplifying:
Average speed = (400 + 40 - 0) / 10
Average speed = 440 / 10
Average speed = 44
Now we need to find the time when the car is going the same speed as its average speed. This means we need to find the time t such that the velocity of the car is equal to the average speed.
Setting the velocity equal to the average speed:
8t + 4 = 44
Solving for t:
8t = 40
t = 5
Therefore, the time when the car is going the same speed as its average speed over the interval 0 to 10 seconds is t = 5. So the answer is C) t = 5.
4. To approximate the y-coordinate when x = 1.01 using the tangent line approximation, we need to find the derivative of the curve equation and then use the derivative to find the slope of the tangent line at the point (1, 1). Finally, we can use the point-slope form of a line to find the y-coordinate.
Let's start by finding the derivative of the curve equation x^2 + sin(xy) + 3y^2 = C with respect to x.
Taking the derivative with respect to x:
2x + cos(xy)(y + xy') + 6yy' = 0
Since we are given that the point (1, 1) lies on the curve, we can substitute these values into our derivative equation:
2(1) + cos(1)(1 + 1y') + 6(1)(y') = 0
Simplifying the equation:
2 + cos(1) + 6y' + cos(1)y' = 0
(cos(1) + 6)y' = -2 - cos(1)
y' = (-2 - cos(1)) / (cos(1) + 6)
Now we have the slope of the tangent line at the point (1, 1). To find the y-coordinate when x = 1.01, we can use the point-slope form of a line:
y - 1 = m(x - 1)
y - 1 = y'(x - 1)
Substituting the values:
y - 1 = [(-2 - cos(1)) / (cos(1) + 6)](1.01 - 1)
Simplifying:
y - 1 = [(-2 - cos(1)) / (cos(1) + 6)](0.01)
y - 1 ≈ -0.003342
Adding 1 to both sides:
y ≈ 0.996658
Therefore, the approximate y-coordinate when x = 1.01 is 0.996. So the answer is A) 0.996.
5. To find the height of the box that uses the least amount of material, we need to determine the relationship between the volume of the box and its dimensions. Let's denote the side length of the square base as x and the height as h.
Given that the volume of the box is 245 cubic centimeters, we can write the volume equation as:
V = x^2 * h = 245
To minimize the amount of material used, we can express h in terms of x using the volume equation:
h = 245 / x^2
Now, we need to find the height that minimizes the surface area of the box. The surface area of the box can be expressed as the sum of the area of the base (x^2) and the sum of the four side areas (4xh):
Surface Area = x^2 + 4xh
Substituting the expression for h:
Surface Area = x^2 + 4x(245 / x^2)
Surface Area = x^2 + 980 / x
To find the height that minimizes the surface area, we can take the derivative of the Surface Area equation with respect to x and set it equal to zero:
d(Surface Area) / dx = (2x - 980 / x^2) = 0
Simplifying:
2x - 980 / x^2 = 0
Multiplying through by x^2 to eliminate the fraction:
2x^3 - 980 = 0
Dividing through by 2:
x^3 - 490 = 0
Solving for x:
x^3 = 490
x ≈ 7.443
Since the question asks for the height in centimeters, we can substitute this value of x back into the equation for h to find the corresponding height:
h = 245 / (7.443)^2
h ≈ 3.942
Therefore, the height in centimeters that minimizes the amount of material used is approximately 3.942 cm. So the answer is C) 3.942 cm.