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
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
3 answers
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.
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!
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!