WATER FLOWS FROM A TANK OF CONSTANT CROSS-SECTIONAL AREA 50 [tex]FT ^2[/tex] THROUGH AN ORIFICE OF CONSTANT CROSS-SECTIONAL AREA [tex]\frac{ 1 }{ 4 }[/tex] [tex]FT ^2[/tex] LOCATED AT THE BOTTOM OF THE TANK. INITIALLY, THE HEIGHT OF THE WATER IN THE TANK WAS 20 FT, AND [tex]T[/tex] SEC LATER IT WAS GIVEN BY THE EQUATION
[tex]2 \sqrt{h}+\frac{1}{25} t-2 \sqrt{20}=0 \quad 0 \leq t \leq 50 \sqrt{20}[/tex]
HOW FAST WAS THE HEIGHT OF THE WATER DECREASING WHEN ITS HEIGHT WAS 3 FT?
A. [tex]100 \sqrt{5}+50 \sqrt{3}[/tex] FT/SEC
B. [tex]\frac{\sqrt{3}}{25}[/tex] FT/SEC
C. [tex]100 \sqrt{5}-50 \sqrt{3} FT / SEC[/tex]
D. [tex]\frac{3}{25}[/tex] FT/SEC
Answer (2)
Differentiate the given equation 2 h + 25 1 t − 2 20 = 0 with respect to time t , obtaining d t d h = − 25 h .
Substitute h = 3 into the expression for d t d h , resulting in d t d h = − 25 3 .
Take the absolute value to find the rate of decrease: d t d h = 25 3 .
The height of the water is decreasing at a rate of 25 3 ft/sec.
Explanation
Problem Setup We are given the equation 2 h + 25 1 t − 2 20 = 0 , which relates the height of the water h to the time t . We want to find how fast the height of the water is decreasing when h = 3 ft. This means we need to find d t d h when h = 3 .
Differentiating the Equation First, we differentiate the given equation with respect to t using implicit differentiation: d t d ( 2 h + 25 1 t − 2 20 ) = d t d ( 0 ) 2 ⋅ 2 h 1 ⋅ d t d h + 25 1 = 0 h 1 d t d h = − 25 1 d t d h = − 25 h
Substituting h = 3 Now, we substitute h = 3 into the expression for d t d h :
d t d h = − 25 3 This means that when the height is 3 ft, the rate of change of the height with respect to time is − 25 3 ft/sec.
Finding the Rate of Decrease Since we want to find how fast the height is decreasing, we take the absolute value of d t d h :
d t d h = − 25 3 = 25 3
Final Answer Therefore, the height of the water is decreasing at a rate of 25 3 ft/sec when its height is 3 ft.
Examples
Imagine you're managing a reservoir and need to predict how quickly the water level drops based on an outflow rate. By understanding the principles of related rates, you can model the water level's change over time. This helps in making informed decisions about water usage and conservation, ensuring sustainable water management for the community.
The height of the water decreased at a rate of 25 3 ft/sec when it was at 3 ft. This was determined by differentiating the given water height equation with respect to time and substituting h = 3 . The correct answer is 25 3 ft/sec.
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