A Receptacle is in the shape of an inverted square pyramid 10cm in height and with a 6cm x 6cm square base. The volume of such pyramid is given by 1/3(area of the base)(height).
Suppose that the receptacle is being filled with water at the rate of .2 cubic centimeters per second. How fast is the water rising when it is 2cm deep?
Any help at all is greatly appreciated!!
2007-10-30 11:49:44 · 1 answers · asked by nicholas_obrien 1 in Science & Mathematics ➔ Mathematics
See diagram:
http://i14.tinypic.com/5xym0hw.gif
When the height and the side of the base of the "pyramid" of water within the receptacle are h and s respectively, the volume of water is
V = s²h/3
From the diagram, s/h = 6/10 (similar triangles). Solve this for s in terms of h and put it into the volume expression. Then differentiate with respect to t. You're told that dV/dt = 0.2 and you are asked for dh/dt when h=2.
2007-10-30 14:19:50 · answer #1 · answered by Ron W 7 · 0⤊ 0⤋