Pregunta
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To test an individual's use of a certain mineral, a researcher injects a small amount of a radioactive form of that mineral into the person's bloodstream. The mineral remaining in the bloodstream is measured each day for several days. Suppose the amount of the mineral remaining in the bloodstream (in miligrams per cubic centimeter) t days after the initial injection is approximated by \( \mathrm{C}(\mathrm{t})=\frac{1}{2}(4 \mathrm{t}+1)^{-1 / 2} \). Find and interpret the rate of change of the mineral level with respect to time for (a) 0 days, (b) 3 days, and (c) 5.5 days. (d) Are the slopes of the tangent lines to C always positive or always negative? What does this indicate about the rate of change of the minerall level? (a) The rate of change of the mineral level with respect to time for 0 days is -1 milligrams per cubic centimeter per day (Round to two decimal places as needed.)

Ask by Alexander Mullins. in the United States
Nov 07,2024

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(a) 0天时变化率为 1 毫克每立方厘米每天。 (b) 3天时变化率为 0.028 毫克每立方厘米每天。 (c) 5.5天时变化率为 0.004 毫克每立方厘米每天。 (d) 斜率总是正的,表示矿物水平随时间增加。

Solución

为了求解这个问题,我们需要计算函数 \( C(t) = \frac{1}{2}(4t + 1)^{-1/2} \) 在特定时间点的导数,即 \( C'(t) \)。导数 \( C'(t) \) 表示的是矿物水平随时间变化的速率。 ### 步骤 1:求导数 \( C'(t) \) 首先,我们需要对函数 \( C(t) = \frac{1}{2}(4t + 1)^{-1/2} \) 求导。 \[ C(t) = \frac{1}{2}(4t + 1)^{-1/2} \] 使用链式法则进行求导: \[ C'(t) = \frac{1}{2} \cdot \left( (4t + 1)^{-1/2} \right)' = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) (4t + 1)^{-3/2} \cdot 4 \] 简化后得到: \[ C'(t) = -\frac{1}{2} \cdot \left( -\frac{1}{2} \right) \cdot 4 (4t + 1)^{-3/2} = \frac{1}{2} \cdot 2 (4t + 1)^{-3/2} = (4t + 1)^{-3/2} \] ### 步骤 2:计算特定时间点的导数值 #### (a) 当 \( t = 0 \) 时 \[ C'(0) = (4 \cdot 0 + 1)^{-3/2} = 1^{-3/2} = 1 \] 所以,0天时矿物水平的变化率是 1 毫克每立方厘米每天。 #### (b) 当 \( t = 3 \) 时 \[ C'(3) = (4 \cdot 3 + 1)^{-3/2} = 13^{-3/2} \] 计算 \( 13^{-3/2} \): \[ 13^{-3/2} \approx 0.028 \] 所以,3天时矿物水平的变化率是 0.028 毫克每立方厘米每天。 #### (c) 当 \( t = 5.5 \) 时 \[ C'(5.5) = (4 \cdot 5.5 + 1)^{-3/2} = 23^{-3/2} \] 计算 \( 23^{-3/2} \): \[ 23^{-3/2} \approx 0.004 \] 所以,5.5天时矿物水平的变化率是 0.004 毫克每立方厘米每天。 ### 步骤 3:分析斜率 #### (d) 斜率的正负 从导数 \( C'(t) = (4t + 1)^{-3/2} \) 可以看出,对于所有 \( t \) 的值,\( (4t + 1) \) 总是正的,因此 \( (4t + 1)^{-3/2} \) 也总是正的。 所以,斜率总是正的,这表明矿物水平随时间增加而增加。 ### 总结 (a) 0天时矿物水平的变化率是 1 毫克每立方厘米每天。 (b) 3天时矿物水平的变化率是 0.028 毫克每立方厘米每天。 (c) 5.5天时矿物水平的变化率是 0.004 毫克每立方厘米每天。 (d) 斜率总是正的,这表明矿物水平随时间增加而增加。

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To evaluate the rate of change of the mineral level with respect to time, we first need to find the derivative of the function \( \mathrm{C}(t) \). Using the formula provided: \[ \mathrm{C}(t) = \frac{1}{2}(4t + 1)^{-1/2} \] We differentiate this using the chain rule: \[ C'(t) = \frac{d}{dt} \left(\frac{1}{2}(4t + 1)^{-1/2}\right) = \frac{1}{2} \cdot -\frac{1}{2}(4t + 1)^{-3/2} \cdot 4 = -\frac{2}{(4t + 1)^{3/2}} \] Now, we can compute the value of the derivative \( C'(t) \) at the specified times: (a) For \( t = 0 \): \[ C'(0) = -\frac{2}{(4(0) + 1)^{3/2}} = -\frac{2}{1^{3/2}} = -2 \text{ mg/cm}^3/\text{day} \] (b) For \( t = 3 \): \[ C'(3) = -\frac{2}{(4(3) + 1)^{3/2}} = -\frac{2}{13^{3/2}} \approx -0.554 \text{ mg/cm}^3/\text{day} \] (c) For \( t = 5.5 \): \[ C'(5.5) = -\frac{2}{(4(5.5) + 1)^{3/2}} = -\frac{2}{23^{3/2}} \approx -0.424 \text{ mg/cm}^3/\text{day} \] (d) The slopes of the tangent lines to \( C(t) \) are always negative, as evidenced by our calculations; this indicates that the rate of change of the mineral level is continually decreasing over time. That means as time progresses, the amount of the mineral in the bloodstream diminishes. Now that we have the results, let’s summarize them: - (a) \( -2 \) mg/cm³/day - (b) \( \approx -0.55 \) mg/cm³/day - (c) \( \approx -0.42 \) mg/cm³/day The general trend of the mineral level in the bloodstream is declining, which is typical for substances that are eventually eliminated from the body.

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