Using Torricelli's Principle, it can be shown that the depth d of a liquid in a bottle with a hole of area 0.5 cm2 in its side can be approximated by
d = 0.0034t2 − 0.52518t + 20,
where t is the time since a stopper was removed from the hole. When will the depth be 13 cm? Round to the nearest tenth of a second
Substitute d with 13. Then solve the resulting quadratic equation.
13 = 0.0034t^2 - 0.52518t + 20
The depth d of a liquid in a bottle with a hole of area 0.5 cm2 in its side can be approximated by
d = 0.0034t2 − 0.52518t + 20,
where t is the time since a stopper was removed from the hole. When will the depth be 10 cm? Round to the nearest tenth of a second.
To find when the depth of the liquid in the bottle will be 13 cm, we can substitute the value of d into the given equation and solve for t.
Given:
d = 13 cm
Substituting the given value into the equation:
13 = 0.0034t^2 - 0.52518t + 20
Rearranging the equation to form a quadratic equation:
0.0034t^2 - 0.52518t + 7 = 0
Now, we can solve this quadratic equation to find the value of t.
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Substituting the values of a, b, and c:
t = (-(-0.52518) ± sqrt((-0.52518)^2 - 4(0.0034)(7))) / (2 * 0.0034)
Simplifying the equation:
t = (0.52518 ± sqrt(0.27577 + 0.0952)) / 0.0068
t = (0.52518 ± sqrt(0.37097)) / 0.0068
Calculating the square root:
t = (0.52518 ± 0.60803) / 0.0068
Now, we will calculate the two possible values of t by adding and subtracting the square root value:
t1 = (0.52518 + 0.60803) / 0.0068
t1 = 112.97755 / 0.0068
t1 ≈ 16616.4 seconds
t2 = (0.52518 - 0.60803) / 0.0068
t2 = -0.08285 / 0.0068
t2 ≈ -12.2 seconds
Since time cannot be negative, we will discard the negative value of t.
Therefore, the time since the stopper was removed when the depth is approximately 13 cm is 16616.4 seconds (rounded to the nearest tenth of a second).
To find the time when the depth of the liquid is 13 cm, we can set up the equation:
13 = 0.0034t^2 - 0.52518t + 20
Now, we need to solve this quadratic equation for t. We can do this by rearranging the equation and setting it equal to zero:
0.0034t^2 - 0.52518t + 7 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 0.0034, b = -0.52518, and c = 7. Plugging these values into the quadratic formula, we get:
t = (-(-0.52518) ± √((-0.52518)^2 - 4(0.0034)(7))) / (2(0.0034))
Simplifying further, we have:
t = (0.52518 ± √(0.2753334028 - 0.0948)) / 0.0068
t = (0.52518 ± √(0.1805334028)) / 0.0068
t = (0.52518 ± 0.425214) / 0.0068
Now, we can calculate two possible values for t:
t₁ = (0.52518 + 0.425214) / 0.0068 ≈ 114.54
t₂ = (0.52518 - 0.425214) / 0.0068 ≈ 6.22
Since time cannot be negative, the solution t₂ = 6.22 is the only valid solution.
Therefore, the depth will be 13 cm approximately 6.22 seconds after the stopper was removed.