answer:Let V be the volume of the water in the cylindrical container, h be the height of the water, and r be the radius of the container. We are given that r = 5 cm and the container is being filled at a rate of dV/dt = 2 cm³/s. The volume of a cylinder is given by the formula V = πr²h. Since the radius of the container is constant, we can write the volume of the water as V = π(5 cm)²h = 25πh. Now, we need to find the rate at which the water level is rising, which is dh/dt. To do this, we'll differentiate both sides of the volume equation with respect to time t: dV/dt = d(25πh)/dt We know dV/dt = 2 cm³/s, so: 2 = 25π(dh/dt) Now, we need to find dh/dt when the height of the water is 4 cm: 2 = 25π(dh/dt) when h = 4 cm To solve for dh/dt, divide both sides by 25π: dh/dt = 2 / (25π) dh/dt ≈ 0.02546 cm/s So, when the height of the water is 4 cm, the water level is rising at a rate of approximately 0.02546 cm/s., To find the rate at which the water level is rising, we first need to determine the cross-sectional area of the cylinder at the level of 4 cm. The height of the water forms a smaller cylinder within the larger one. The radius of this smaller cylinder can be found using the Pythagorean theorem: Radius of smaller cylinder = √(Radius of larger cylinder^2 - Height of water^2) = √(5^2 - 4^2) = √(25 - 16) = √9 = 3 cm Now, let's calculate the cross-sectional area A of the smaller cylinder: A = π * r^2 = π * (3 cm)^2 = 9π cm^2 We know that the volume V of the smaller cylinder being filled is increasing at a rate of 2 cm^3/s. We can now relate the rate of change of volume to the rate of change of height h using the following equation: dV/dt = A * (dh/dt) Where dV/dt is the rate of change of volume (2 cm^3/s), A is the cross-sectional area of the smaller cylinder (9π cm^2), and dh/dt is the rate of change of height (which we want to find). Rearranging the equation to solve for dh/dt, we get: dh/dt = (dV/dt) / A ≈ (2 cm^3/s) / (9π cm^2) ≈ 0.0698 cm/s (rounded to three decimal places) So, the water level is rising at a rate of approximately 0.0698 cm/s when the height of water is 4 cm.,To find the rate at which the water level is rising, we can use the formula: Volume of water = π * r^2 * h where r is the radius (in this case, 5 cm), h is the height of the water (in this case, 4 cm), and the volume is being filled at a rate of 2 cm³/sec. When the height (h) of the water is 4 cm, the volume is: V = π * (5 cm)^2 * 4 cm V = π * 25 cm² * 4 cm V = 100π cm³ Since the volume is being filled at a rate of 2 cm³/sec: 2 cm³/sec = (2 cm³/sec) / (100π cm³) To find the rate at which the water level is rising (dh/dt) when h = 4 cm: dh/dt = (2 cm³/sec) / (100π cm³) dh/dt = 2/100π cm/sec dh/dt = 1/(50π) cm/sec So, the water level is rising at a rate of 1/(50π) cm/sec, or approximately 0.00549 cm/sec when the height of water is 4 cm.

answer:To solve this problem, we can use the Pythagorean theorem to relate the length of the ladder, the distance from the base to the wall, and the distance from the top of the ladder to the ground. Let x be the distance from the base of the ladder to the wall, and y be the distance from the top of the ladder to the ground. We are given that the ladder has a length of 10 meters and initially x = 6 meters. According to the Pythagorean theorem, x^2 + y^2 = 10^2 When the base is 4 meters away from the wall (x = 4), we want to find the rate at which the top of the ladder is moving down the wall (dy/dt). First, find the value of y when x = 4: (4)^2 + y^2 = 100 16 + y^2 = 100 y^2 = 84 y = √84 (approximately 9.1667 meters) Now, we can use implicit differentiation on the equation: d/dt (x^2 + y^2) = d/dt (100) 2x(dx/dt) + 2y(dy/dt) = 0 Plug in the given values when x = 4 and y = √84: 2(4)(dx/dt) + 2(√84)(dy/dt) = 0 We know that the ladder is being pushed towards the wall at a rate of 0.2 meters per second, so: dx/dt = -0.2 Substitute this value into the equation: 2(4)(-0.2) + 2(√84)(dy/dt) = 0 Solve for dy/dt: -1.6 + 2(√84)(dy/dt) = 0 2(√84)(dy/dt) = 1.6 dy/dt = 1.6 / (2(√84)) dy/dt ≈ -0.2343 meters per second (rounded to four decimal places) So, when the base,Let x be the distance of the base of the ladder from the wall, and y be the distance of the top of the ladder from the ground. We are given that the ladder has a length of 10 meters, so we can use the Pythagorean theorem to relate x and y: x^2 + y^2 = 10^2 = 100 We are also given that the base of the ladder is being pushed towards the wall at a rate of 0.2 meters per second, so dx/dt = -0.2 m/s (negative because the distance is decreasing). We want to find dy/dt when x = 4 meters. To find dy/dt, we can differentiate the equation with respect to time t: d(x^2)/dt + d(y^2)/dt = d(100)/dt 2x(dx/dt) + 2y(dy/dt) = 0 Now we can plug in the given values and solve for dy/dt: 2(4)(-0.2) + 2y(dy/dt) = 0 When x = 4, we can find y using the Pythagorean theorem: 4^2 + y^2 = 100 y^2 = 100 - 16 = 84 y = √84 ≈ 9.17 Now we can plug in the value of y and solve for dy/dt: 2(4)(-0.2) + 2(9.17)(dy/dt) = 0 -1.6 + 18.34(dy/dt) = 0 Now, solve for dy/dt: 18.34(dy/dt) = 1.6 dy/dt = 1.6 / 18.34 ≈ 0.0872 m/s The top of the ladder is moving down the wall at a rate of approximately 0.09 meters per second when the base is 4 meters away from the wall., The top of the ladder is moving down the wall at 0.44 m/s when the base is 4 meters away from the wall. Here's the reasoning: 1. Let x be the distance of the base of the ladder from the wall, and y be the height of the top of the ladder on the wall. 2. We have the equation for a right triangle: x^2 + y^2 = 10^2 3. Differentiating both sides with respect to time (t), we get: 2x(dx/dt) + 2y(dy/dt) = 0 4. When x = 4 meters, y = sqrt(100-16) = 8.06 meters (rounded to two decimal places). 5. Now we can find dy/dt when dx/dt = 0.2 m/s and x = 4 meters: dy/dt = -(2x/2y) * dx/dt = -(4/8.06) * 0.2 ≈ -0.44 m/s Since the value of dy/dt is negative, it means that the top of the ladder is moving downward. Therefore, the top of the ladder is moving down the wall at 0.44 m/s when the base is 4 meters away from the wall.

answer:To find the rate at which the water level is rising, we can use the formula: Rate of change in volume = Volume / Time taken First, let's find the volume of the water tank at the given depth of 1 meter. The volume of a rectangular prism (including a tank) is given by the formula: Volume = Length * Width * Height At a depth of 1 meter, the tank will contain water with a volume: Volume = 3m * 2m * 1m = 6 cubic meters Now, let's find the time taken for the water level to reach 1 meter. Since water is being pumped into the tank at a rate of 0.5 cubic meters per minute, we can find the time taken by setting up an equation: Volume = Pumping rate * Time Solving for Time, we have: Time = Volume / Pumping rate = 6 cubic meters / 0.5 cubic meters per minute = 12 minutes Now, let's find the rate of change in volume: Rate of change in volume = Volume / Time = 6 cubic meters / 12 minutes = 0.5 cubic meters / minute So, the water level is rising at a rate of 0.5 cubic meters per minute when the depth of the water is 1 meter.,To find the rate at which the water level is rising, we can use the formula: Rate of water level rise = (Rate of water being pumped in) / (Area of the base of the tank) First, let's find the area of the base of the tank. Since the tank has a rectangular shape, the area of the base can be found by multiplying the length and the width: Area of the base = Length × Width Area of the base = 3 meters × 2 meters Area of the base = 6 square meters Now, we can find the rate at which the water level is rising: Rate of water level rise = (Rate of water being pumped in) / (Area of the base of the tank) Rate of water level rise = (0.5 cubic meters per minute) / (6 square meters) Rate of water level rise = 0.5 / 6 Rate of water level rise ≈ 0.0833 meters per minute So, when the depth of the water is 1 meter, the water level is rising at a rate of approximately 0.08 meters per minute (rounded to the nearest hundredth)., First, we need to find the volume of the water in the tank when the depth is 1 meter. We can do this by calculating the volume of the rectangular prism that represents the tank below the water level. Volume = Length * Width * Height = 3 m * 2 m * 1 m (since the height here is the depth of the water) = 6 cubic meters Now, let's determine how long it takes to fill this volume at the given flow rate: Time = Volume / Flow rate = 6 m³ / 0.5 m³/min = 12 minutes Next, we need to find the cross-sectional area of the tank at the water level of 1 meter: Area = Length * Width = 3 m * 2 m = 6 square meters Finally, we can calculate the rate of change in the water level: Rate = Area / Time = 6 m² / 12 min = 0.5 m/min So, the water level in the tank is rising at a rate of 0.5 meters per minute when the depth of the water is 1 meter.

answer:To find the rate at which the water level is rising, we first need to find the radius of the cone at a height of 6 feet. We can do this by using the similar triangles property of a cone. The ratio of the height to the radius is constant for all points on the cone. So, let's set up the proportion: Height (10 ft) / Radius (5 ft) = Height (6 ft) / Radius (x) Cross-multiply and solve for x: 10x = 5 * 6 x = 3 feet Now that we have the radius when the water level is 6 feet deep, let's set up our related rates equation. The volume of a cone is given by: Volume = (1/3)πr²h We want to find dV/dt, the rate at which the volume is changing with respect to time. Using implicit differentiation, we get: dV/dt = (1/3) * 2πr * (dr/dt) * h + (1/3) * πr² * (dh/dt) ,To solve this problem, we will use the related rates equation for the volume of a cone. The volume V of a cone with radius r and height h is given by the formula: V = (1/3)πr^2h We are given that the tank is being filled at a rate of 5 cubic feet per minute, so dV/dt = 5 ft³/min. We want to find the rate at which the water level is rising, which is dh/dt, when the water is 6 feet deep (h = 6). Since the tank has a fixed shape, the ratio of the radius to the height is constant: r/h = 5/10 = 1/2 So, r = (1/2)h. Now, we will differentiate the volume equation with respect to time t: dV/dt = d/dt(1/3)πr^2h Using the chain rule, we get: 5 = (1/3)π(2r(dr/dt)h + r^2(dh/dt)) Now, we will substitute r = (1/2)h: 5 = (1/3)π((1/2)h)^2(dh/dt) + (1/2)h(2(1/2)h(dh/dt)) When the water is 6 feet deep (h = 6), we can find the corresponding radius: r = (1/2)(6) = 3 Now, we will substitute h = 6 and r = 3 into the equation: 5 = (1/3)π(3^2(dh/dt) + 3(6)(dh/dt)) 5 = (1/3)π(9(dh/dt) + 18(dh/dt)) Now, we will solve for dh/dt: 5 = (1/3)π(27(dh/dt)) 15 = π(27(dh/dt)) 15/(27π) = dh/dt dh/dt = 5/(9π) So, the rate at which the water level is rising when the water is 6 feet deep is 5/(9π) feet per minute.,We'll use the formula for the volume of a cone in terms of its radius and height: V = frac{1}{3} pi r^2 h where r is the radius and h is the height of the cone. As the tank is being filled, the height of water h is increasing. We'll denote it as h(t), where t is the time in minutes. Since the radius remains constant, we will set the radius to r = 5 feet. The rate at which the water level is rising is given by: frac{dh}{dt}. Using the related rates equation, we know that the rate at which the volume is increasing as the tank fills is: dV/dt = 5 ft³/min. Then, we differentiate the volume equation with respect to time: frac{dV}{dt} = frac{1}{3} pi (2r) cdot frac{dr}{dt} + frac{1}{3} pi r^2 cdot frac{dh}{dt} = 5 implies boxed{5 + frac{1}{3} pi cdot 25 cdot frac{dh}{dt} = 5} Solving for the rate at which the water level is rising frac{dh}{dt}: frac{dh}{dt} = frac{1}{125 pi} = frac{1}{376.991} ft/min At h = 6text{ ft}, the rate at which the water level is rising is: frac{dh}{dt} = frac{1}{376.991} text{ ft/min}