Calculus

what is the solution for this?ladder 20 ft. long is leaning against an embankment inclined 60degrees to the horizontal. If the bottom of the ladder is beingmoved horizontally toward the embankment at 1 ft/sec,how fast isthe top of the ladder moving when the bottom is 4ft from theembankment?

i dont know how to get to the solution but the answer is:(1/194)*(3sqrt(97)+97) ft/sec or .65 feet per second. EXPERT ANSWER AC2 = AB2 + BC2 =Y2 + X2 = 400. Hence 2Y(dY/dt) + 2X(dX/dt) =0. …..(1),When BC = 4, AB = √(400-42) =√(400-16) = √(384). Thus when X = 4, Y =√(384)Substitute in (1) to get 2√(384)(dY/dt) + 2 …

what is the solution for this?ladder 20 ft. long is leaning against an embankment inclined 60degrees to the horizontal. If the bottom of the ladder is beingmoved horizontally toward the embankment at 1 ft/sec,how fast isthe top of the ladder moving when the bottom is 4ft from theembankment? Read More »

A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate 2m^3 per min, find the rate at which the water level is rising when the water is 3m deep.1. Sketch a picture with depicting known parameters.2. Find the formula for the volume of the water in the tank.

3. To eliminate one variable, express r as a function of h. 4. Write the formula for the volume of the water as a function of height h. 5. Write down the known and unknown rates. 6. Write the equation relating the two rates. 7. Solve for the unknown rate and answer the question. EXPERT …

A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate 2m^3 per min, find the rate at which the water level is rising when the water is 3m deep.1. Sketch a picture with depicting known parameters.2. Find the formula for the volume of the water in the tank. Read More »

Find integral_0^2 (2x^3 – 6x + 2/x^2 + 1) dx and interpret the result in terms of areas. The Fundamental Theorem gives integral_0^2 (2x^3 – 6x + 2/x^2 + 1) dx = 2(x^4/4) – 6(x^2/2) + 2 tan^-1(x)]_0^2 This is the exact value of the integral. If a decimal approximation is desired, we can use a calculator to approximate tan^-1(2). Doing so, we get integral_0^2 (2x^3 – 6x + 2/x^2 + 1) dx (Round your answer to four decimal places.) The figure below shows the graph of the integrand. We know that the value of the integral can be interpreted as a net area: the sum of the areas labeled with a plus sign minus the area labeled with a minus sign.

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