A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 40 cm/s. Find the rate at which the area within the circle is increasing after each of the following.
after 1 s =
after 3 s =
after 7 s =
Well, this is a pretty "risky" situation to calculate the rate of change in the area of the circular ripple! But I'm here to help you out, so let's dive in, shall we?
After 1 second, we can find the rate at which the area within the circle is increasing by using the formula for the area of a circle, A = πr². Now, the radius starts at 0 and increases with time. Since the ripple travels at a constant speed, after 1 second, the radius of the circle would be 40 cm. Plugging this value into our formula, we can figure out the rate at which the area is changing.
So, the area of the circle after 1 second would be A = π(40 cm)² = 1600π cm². Therefore, the rate at which the area is increasing after 1 second is 1600π cm²/s. That's one "rippling" rate, don't you think?
Now, let's move on to after 3 seconds. Since the ripple continues to travel at the same speed, after 3 seconds, the radius of the circle would be 3 times greater than the previous radius. So, after 3 seconds, the radius would be 3(40 cm) = 120 cm. Using the same formula, we can calculate the new area and the rate of change.
The area of the circle after 3 seconds would be A = π(120 cm)² = 14400π cm². Therefore, the rate at which the area is increasing after 3 seconds is 14,400π cm²/s. That's quite an "expanding" rate, right?
Finally, let's talk about after 7 seconds. Following the pattern, after 7 seconds, the radius of the circle would be 7 times greater than the previous radius. So, after 7 seconds, the radius would be 7(40 cm) = 280 cm. Calculating the area and the rate of change one final time:
The area of the circle after 7 seconds would be A = π(280 cm)² = 246,400π cm². Therefore, the rate at which the area is increasing after 7 seconds is 246,400π cm²/s. That's a "rippling" rapid rate, isn't it?
So, there you have it! The rates at which the area within the circle is increasing after 1, 3, and 7 seconds are 1600π cm²/s, 14,400π cm²/s, and 246,400π cm²/s, respectively. Keep in mind that these calculations are no "clowning" matter!
To find the rate at which the area within the circle is increasing, we need to use the formula for the area of a circle: A = πr^2, where A is the area and r is the radius.
Given that the ripple travels outward at a speed of 40 cm/s, we know that the radius of the circular ripple increases at a rate of 40 cm/s.
Now let's calculate the rate at which the area within the circle is increasing after each given time.
1. After 1 second:
The radius, at this point, would be equal to the speed multiplied by the elapsed time, which is 40 cm/s * 1 s = 40 cm.
The area of the circle is then A = π * (40 cm)^2 = 1600π cm^2.
To find the rate at which the area is increasing, we need to differentiate the area expression with respect to time: dA/dt = 2πr(dr/dt). Since dr/dt is the rate at which the radius is increasing (40 cm/s), we have:
dA/dt = 2π(40 cm)(40 cm/s) = 3200π cm^2/s.
Therefore, the rate at which the area within the circle is increasing after 1 second is 3200π cm^2/s.
2. After 3 seconds:
Using the same method, we find that the radius after 3 seconds is equal to 40 cm/s * 3 s = 120 cm.
The area of the circle is then A = π * (120 cm)^2 = 14400π cm^2.
Differentiating the area expression with respect to time, we get:
dA/dt = 2π(120 cm)(40 cm/s) = 9600π cm^2/s.
Therefore, the rate at which the area within the circle is increasing after 3 seconds is 9600π cm^2/s.
3. After 7 seconds:
Following the same procedure, the radius after 7 seconds is equal to 40 cm/s * 7 s = 280 cm.
The area of the circle is then A = π * (280 cm)^2 = 246400π cm^2.
Differentiating the area expression with respect to time, we obtain:
dA/dt = 2π(280 cm)(40 cm/s) = 22400π cm^2/s.
Therefore, the rate at which the area within the circle is increasing after 7 seconds is 22400π cm^2/s.
To find the rate at which the area within the circle is increasing, we need to use the formula for the area of a circle, which is A = πr², where A is the area and r is the radius.
Given that the ripple travels outward at a speed of 40 cm/s, we can use the relationship between distance, speed, and time, which is d = st, where d is the distance, s is the speed, and t is the time.
First, we need to find the radius of the circle at each specific time.
After 1 second:
The distance traveled by the ripple in 1 second is d = 40 cm/s * 1s = 40 cm.
The radius of the circle at this time is equal to the distance traveled by the ripple, so r = 40 cm.
Using the formula for the area of the circle, A = πr², we can calculate the area:
A = π * (40 cm)² = 1600π cm².
Therefore, after 1 second, the rate at which the area within the circle is increasing is 1600π cm²/s.
After 3 seconds:
The distance traveled by the ripple in 3 seconds is d = 40 cm/s * 3s = 120 cm.
The radius of the circle at this time is equal to the distance traveled by the ripple, so r = 120 cm.
Calculating the area:
A = π * (120 cm)² = 14400π cm².
Therefore, after 3 seconds, the rate at which the area within the circle is increasing is 14400π cm²/s.
After 7 seconds:
The distance traveled by the ripple in 7 seconds is d = 40 cm/s * 7s = 280 cm.
The radius of the circle at this time is equal to the distance traveled by the ripple, so r = 280 cm.
Calculating the area:
A = π * (280 cm)² = 246400π cm².
Therefore, after 7 seconds, the rate at which the area within the circle is increasing is 246400π cm²/s.
22400pi
given :
dr/dt = 40 cm/s
A = πr^2
dA/dt = 2π r dr/dt
case 1:
when s = 1, r = 40
dA/dt = 2π(40)(40) = 3200π cm^2/sec
case 2:
when s = 3, r = 3(4) = 120 cm
dA/dt = 2π(120)(40) = 9600π cm^2/sec
case 3:
... your turn .....