A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top). If water is poured into the cup at a rate of 2 cm^{3}/s, how fast is the water level rising when the water is 5 cm deep?

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Jeffrey Jordon · Accepted Answer

Step 1If the water level is x and the radius of the cone at this height is&amp;nbsp;rx, then we can use the properties of similar triangles to writerx3=x10Multiply both sides by 3rx=0.3xThe volume of the water in the cone is given byV=13πr2h=13π(0.3x)2xDifferentiate both sides with respect to tdVdt=0.03πd(x3)dtSince the water is being poured at&amp;nbsp;2cm3/s,dV/dt=2Use chain rule in the right-hand side2=0.03π(3x3−1)⋅dxdtSolve for&amp;nbsp;dx/dtdxdt=20.09πx2=2009πx2When the water is 5 cm deep, we have x=5dxdt|x=5=2009π⋅52≈0.283cm/sResultweter level is rising at the rate of&amp;nbsp;≈0.283cm/s&amp;nbsp;

user_27qwe · Accepted Answer

Step 1 V=volume of water r=radius of water h=heidht of water V=13πr2h Step 2 Find r in terms of h. rh=310 r=3h10 Step 3 Substitute for r. V=13π(3h10)2h V=3100πh3 Step 4 Differentiate dVdt=3100π⋅3h2dhdt Step 5 Substitute given information 2=3100π⋅3(5)2⋅dhdt Step 6 Solve for dhdt 2=9π4⋅dhdt dhdt=89π≈.283 Result 89π≈.283cm/sec

A paper cup has the shape of a cone with

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