Recent content by BFPerkins

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    Rate of change: Height of a conical pile

    thanks for the help. Abput 15 minutes ago. i saw where i made my mistake and finally came up with the corredt answer.
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    Rate of change: Height of a conical pile

    By any chance, has any0ne else worked this problem out and come with an answer of 8/405pi? When I put the height as a function of the radius, I get h = 2/3r^2 so the volume = 2/9\pi h^3 when I differentiate both sides of the equation with respect to t, I get V\prime = 2/9\pi h^2 h\prime...
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    Rate of change: Height of a conical pile

    Since I am making the height as a function of the radius, I don't think I need to squlare the 1.5. Instead, if i just write [(2/3)πh^3]/3, or (2/9)πh^3, I can differentiate thar and hopefully I will get thte corect answer.
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    Rate of change: Height of a conical pile

    Thanks! I see wehere I made my mistake. I went with 2/9\pi\r^3 Iwas eliminatimg the wrong variable. I appreciate the help.
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    Rate of change: Height of a conical pile

    I have that. r Approximatelly equals 1.5h, however, I do not know how to find the rate of change for the radius with respect to time (dr/dt) in order to plug that into my equation and solve for the change in height with respect to time t (dh/dt). Since both radius and height change as the...
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    Rate of change: Height of a conical pile

    The problem states: sand falls onto a conical pile at a gravel yard at a rare of 10 cubic feet per minute. The base of the pile is approximately three times the altitude. How fast is the pile getting taller when the pile is 15 feet tall?Volume = πr² h/3 dV =...
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    Volume of a function around an axis

    I'm back and I've carefully checked to make sure I am reading it correctly. First I set function in terms of y because of its rotation around the verticle axis, which gives me y = sqrt(2x-2) Then I subtracted 3 because it revolves around x = 3. This is my radius, so I square it to get my...
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    Volume of a function around an axis

    Sorry! Never mind the question. I was looking at the problem wrong. The function is bounded by y = 0 and y = 4, not x. Now it is a piece of cake!
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    Volume of a function around an axis

    Given this problem: Find the volume and describe the shape of the object formed by the function f(x) = ½ x²+2 when the function is rotated around the x =3 axis and bounded by th region between the x = 0 and x = 4. I am not sure if I am thihnking about this correctly. I know in order to...
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