1.At noon of a certain day,ship A is ^0 km due north of ship B. If A sails east at 12 km/hr and B sails north 9 km/hr,determine how rapidly the distance between them is changing 2 hrs later. Is it increasing or decreasing?
2.At a given instant the legs of a right triangle are 16 cm and 12 cm, respectively. The first leg decreases at 1/2 cm/min and the second increases at 2 cm/min. At what rate is the area increasing after 2 min?
#1 contains a typo,
fix the ^0
#2
let the first leg be x cm
and the second leg be y cm
given:
dx/dt = -.5
dy/dt = 2
find: dA/dt , when t = 2
A = (1/2)xy
dA/dt = (1/2)(x dy/dt + y dx/dt)
at the given time of 2 minutes
x = 16 - 2(.5) = 15
y = 12 + 2(2) = 16
dA/dt = (1/2)(15(2) + 16(-.5) )
= 11
The area at that instant is increasing at 11 cm^2/min
check my arithmetic, I am only on my first coffee.
At noon of a certain day,ship A is 60 km due north of ship B. If A sails east at 12 km/hr and B sails north 9 km/hr,determine how rapidly the distance between them is changing 2 hrs later. Is it increasing or decreasing?
To determine the rate at which the distance between two moving objects is changing or how rapidly it is changing, we can use the concept of related rates. In both the given problems, we need to find the rate of change of a certain quantity with respect to time.
1. For the first problem, we need to find how rapidly the distance between the two ships is changing after 2 hours. Let's consider the position of the ships at that moment.
At noon, ship A is at a position P1, which is ^0 km due north of ship B. After 2 hours, ship A will have moved east for 2 hours × 12 km/hr = 24 km. Now, let's find the new positions of ship A and ship B.
Ship A's new position P2: ^24 km east, ^0 km north
Ship B's new position P3: ^0 km east, ^18 km north
To determine the distance between the two ships, we can use the Pythagorean theorem:
Distance between A and B = √((x2 - x1)^2 + (y2 - y1)^2)
Distance between A and B = √((0 - 24)^2 + (18 - 0)^2)
Distance between A and B = √((-24)^2 + 18^2)
Distance between A and B = √(576 + 324)
Distance between A and B = √900
Distance between A and B = 30 km
Now, let's find the rate of change of this distance after 2 hours. We can use the concept of related rates by differentiating the distance equation with respect to time.
d(Distance between A and B)/dt = d(√900)/dt
Since A and B are moving, their x and y coordinate values will change over time. However, in this problem, we are only given their rates of change. Using the given velocities, we know the rates of change for x and y coordinates: dx/dt = 12 km/hr for ship A, and dy/dt = 9 km/hr for ship B.
To solve for d(Distance between A and B)/dt, we need to differentiate the distance function with respect to time:
d(Distance between A and B)/dt = (1/2)*(900)^(-1/2)*2(24)(12) + (1/2)*(900)^(-1/2)*2(18)(9)
Simplifying the expression:
d(Distance between A and B)/dt = 24(12)√900 + 18(9)√900
d(Distance between A and B)/dt = 288 + 162
d(Distance between A and B)/dt = 450 km/hr
Therefore, the distance between the two ships is changing at a rate of 450 km/hr after 2 hours. Since this value is positive, it means the distance between them is increasing.
2. For the second problem, we need to find the rate at which the area of the right triangle is increasing after 2 minutes. Let's consider the given triangle with legs measuring 16 cm and 12 cm.
At the given instant, the triangle's legs are 16 cm and 12 cm, respectively. The area of a right triangle is given by the formula: Area = (1/2) × base × height.
In this case, the base of the triangle is the decreasing leg, which is changing at a rate of -1/2 cm/min, and the height is the increasing leg, which is changing at a rate of 2 cm/min.
To find the rate at which the area is increasing, we can differentiate the area function with respect to time:
d(Area)/dt = (1/2) × (d(base)/dt) × height + (1/2) × base × (d(height)/dt)
Substituting the given values:
d(Area)/dt = (1/2) × (-1/2) × 12 + (1/2) × 16 × 2
Simplifying the expression:
d(Area)/dt = -6 + 16
d(Area)/dt = 10 cm^2/min
Therefore, the area of the right triangle is increasing at a rate of 10 cm^2/min after 2 minutes.