Differential Calculus Solutions
Topics || Problems
A triangular trough is 10ft long 4ft across the top and 4 feet deep. If water flows in at the rate of 30 cu. ft. per min., find how fast the surface rising when the water is 6 in. deep.
Solution:
By ratio and proportion on the triangle, at any given height, the base is same as the height (\(b = h\)). Thus at a base of the triangle at \(h = 0.5 ft\) is also \(0.5 ft\)
\(Vol = A_{triangle}(10)\)
\(Vol = \frac{1}{2}bh (10)\)
As the water increases, the value of b and h are also changing, thus, b and h are not constants.
\(Vol = 5 h^2\), Since \(b = h\)
\(\frac{dV}{dt} = 10 h \frac{dh}{dt}\)
At h =0.5 ft\(3 = 10 (0.5) \frac{dh}{dt}\)
\(\frac{dh}{dt} = 0.6 \text{ft per min}\)
Thus the surface is rising at 0.6 feet per minute.