Q#1: From a point of O, two cars leave at the same time. One car travels west and after ‘t’ sec, its position is x=t2+t ft. the other car travels north and it covers y= t2+3t ft in ‘t’ sec. At what rate is the distance between the two cars changing after 5 sec?
Question#2: Sand falls from a container at the rate of 10 ft3/min and forms a conical pile whose height is always double the radius of the base. How fast is the height increasing when the pile is 5 ft high?
To find the rate at which the distance between the two cars is changing after 5 seconds, we need to use the concept of derivatives.
1. First, let's find the positions of the two cars as a function of time:
- For the car traveling west, we are given the position function x = t^2 + t.
- For the car traveling north, we are given the position function y = t^2 + 3t.
2. The distance between the two cars can be found using the distance formula:
d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]
3. Let's substitute the positions of the two cars into the distance formula:
d = sqrt[(t^2 + t - 0)^2 + (t^2 + 3t - 0)^2]
d = sqrt[(t^4 + 2t^3 + t^2) + (t^4 + 6t^3 + 9t^2)]
4. Now, let's differentiate the distance function with respect to time (t):
dd/dt = d/dt[sqrt[(t^4 + 2t^3 + t^2) + (t^4 + 6t^3 + 9t^2)]]
5. Simplify the derivative by applying the chain rule and power rule:
dd/dt = [(4t^3 + 6t^2 + 2t)(t^4 + 6t^3 + 9t^2) + (4t^3 + 2t^2 + 2t)(t^4 + 2t^3 + t^2)] / (2sqrt[(t^4 + 2t^3 + t^2) + (t^4 + 6t^3 + 9t^2)])
6. Plug in t = 5 into the derivative to get the rate at which the distance is changing after 5 seconds.
To find the rate at which the height of the pile is increasing when the pile is 5 ft high, we also need to use derivatives:
1. Let's denote the height of the pile as h and the radius of the base as r.
- The rate at which the sand is falling is given as 10 ft^3/min.
2. We know the relationship between the height and radius of the conical pile:
h = 2r
3. To relate the height and radius to the volume, we need the formula for the volume of a cone:
V = (1/3)πr^2h
4. We are given that the volume is changing at a rate of 10 ft^3/min:
dV/dt = 10
5. Substitute the relationship between h and r into the volume formula:
V = (1/3)πr^2(2r)
V = (2/3)πr^3
6. Differentiate the volume equation with respect to time (t):
dV/dt = d/dt[(2/3)πr^3]
7. Substitute the given rate of change of volume and solve for dr/dt (the rate of change of radius):
10 = d/dt[(2/3)πr^3]
8. Solve for dr/dt to find the rate at which the height of the pile is increasing when the pile is 5 ft high.