Solved on Feb 02, 2024

(a) Find depth where light intensity is half of surface intensity $L$ . Depth = $D$ meters. (b) Light intensity at 11 m depth = $I = L \cdot e^{(-1.95 \cdot 11)}$ . (c) Find depth where light intensity is 2.4% of surface intensity $L$ . Depth = $D$ meters.

STEP 1

Assumptions
1. The intensity of light at the surface of the lake is $L$ .
2. The intensity of light at a depth of $x$ meters below the surface satisfies the differential equation $\frac{dI}{dx} = (-1.95)I$ .
3. We are looking for the depth at which the intensity is half of $L$ , the intensity at a depth of 11 meters, and the depth at which the intensity is 2.4% of $L$ .

STEP 2

To solve the differential equation, we will use separation of variables. We want to separate the $I$ terms on one side and the $x$ terms on the other side.
$\frac{1}{I}dI = -1.95dx$

STEP 3

Integrate both sides of the equation to find the general solution.
$\int \frac{1}{I}dI = \int -1.95dx$

STEP 4

Perform the integration.
$\ln|I| = -1.95x + C$

STEP 5

Exponentiate both sides to solve for $I$ .
$I = e^{-1.95x + C}$

STEP 6

Since the intensity of light at the surface ( $x=0$ ) is $L$ , we can find the constant $C$ by plugging in $x=0$ and $I=L$ .
$L = e^{C}$

STEP 7

Solve for $C$ .
$C = \ln(L)$

STEP 8

Substitute $C$ back into the equation for $I$ .
$I = e^{-1.95x + \ln(L)}$

STEP 9

Simplify the equation using properties of exponents.
$I = Le^{-1.95x}$

STEP 10

Now we will find the depth at which the intensity is half of $L$ . Set $I = \frac{1}{2}L$ and solve for $x$ .
$\frac{1}{2}L = Le^{-1.95x}$

STEP 11

Divide both sides by $L$ to isolate the exponential term.
$\frac{1}{2} = e^{-1.95x}$

STEP 12

Take the natural logarithm of both sides to solve for $x$ .
$\ln\left(\frac{1}{2}\right) = -1.95x$

STEP 13

Solve for $x$ .
$x = \frac{\ln\left(\frac{1}{2}\right)}{-1.95}$

STEP 14

Calculate the value of $x$ .
$x = \frac{\ln(0.5)}{-1.95}$

STEP 15

Use a calculator to find the numerical value of $x$ .
$x = \frac{-0.6931}{-1.95} \approx 0.3554$

STEP 16

Round the depth to an appropriate number of significant figures, if necessary. Since the problem does not specify, we can leave it as is or round to two decimal places.
$x \approx 0.36 \text{ meters}$

STEP 17

For part (b), we need to find the intensity of light at a depth of 11 meters. Use the equation $I = Le^{-1.95x}$ with $x=11$ .
$I = Le^{-1.95 \cdot 11}$

STEP 18

Calculate the intensity at 11 meters.
$I = Le^{-21.45}$

STEP 19

For part (c), we need to find the depth at which the intensity is 2.4% of that at the surface. Set $I = 0.024L$ and solve for $x$ .
$0.024L = Le^{-1.95x}$

STEP 20

Divide both sides by $L$ to isolate the exponential term.
$0.024 = e^{-1.95x}$

STEP 21

Take the natural logarithm of both sides to solve for $x$ .
$\ln(0.024) = -1.95x$

STEP 22

Solve for $x$ .
$x = \frac{\ln(0.024)}{-1.95}$

STEP 23

Use a calculator to find the numerical value of $x$ .
$x = \frac{\ln(0.024)}{-1.95}$

STEP 24

Calculate the depth to an appropriate number of significant figures, if necessary.
$x \approx \frac{-3.7337}{-1.95} \approx 1.9152$

STEP 25

Round the depth to an appropriate number of significant figures, if necessary. Since the problem does not specify, we can leave it as is or round to two decimal places.
$x \approx 1.92 \text{ meters}$
Solutions: (a) The depth at which the intensity of light is half of $L$ is approximately $0.36$ meters. (b) The intensity of light at a depth of 11 meters is $Le^{-21.45}$ . (c) The depth at which the intensity of light is 2.4 percent of that at the surface is approximately $1.92$ meters.

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(a) Find depth where light intensity is half of surface intensity LLL. Depth = DDD meters. (b) Light intensity at 11 m depth = I=L⋅e(−1.95⋅11)I = L \cdot e^{(-1.95 \cdot 11)}I=L⋅e(−1.95⋅11). (c) Find depth where light intensity is 2.4% of surface intensity LLL. Depth = DDD meters.

STEP 1

STEP 2

To solve the differential equation, we will use separation of variables. We want to separate the III terms on one side and the xxx terms on the other side.1IdI=−1.95dx\frac{1}{I}dI = -1.95dxI1​dI=−1.95dx

STEP 3

Integrate both sides of the equation to find the general solution.∫1IdI=∫−1.95dx\int \frac{1}{I}dI = \int -1.95dx∫I1​dI=∫−1.95dx

STEP 4

Perform the integration.ln⁡∣I∣=−1.95x+C\ln|I| = -1.95x + Cln∣I∣=−1.95x+C

STEP 5

Exponentiate both sides to solve for III.I=e−1.95x+CI = e^{-1.95x + C}I=e−1.95x+C

STEP 6

Since the intensity of light at the surface (x=0x=0x=0) is LLL, we can find the constant CCC by plugging in x=0x=0x=0 and I=LI=LI=L.L=eCL = e^{C}L=eC

STEP 7

Solve for CCC.C=ln⁡(L)C = \ln(L)C=ln(L)

STEP 8

Substitute CCC back into the equation for III.I=e−1.95x+ln⁡(L)I = e^{-1.95x + \ln(L)}I=e−1.95x+ln(L)

STEP 9

Simplify the equation using properties of exponents.I=Le−1.95xI = Le^{-1.95x}I=Le−1.95x

STEP 10

Now we will find the depth at which the intensity is half of LLL. Set I=12LI = \frac{1}{2}LI=21​L and solve for xxx.12L=Le−1.95x\frac{1}{2}L = Le^{-1.95x}21​L=Le−1.95x

STEP 11

Divide both sides by LLL to isolate the exponential term.12=e−1.95x\frac{1}{2} = e^{-1.95x}21​=e−1.95x

STEP 12

Take the natural logarithm of both sides to solve for xxx.ln⁡(12)=−1.95x\ln\left(\frac{1}{2}\right) = -1.95xln(21​)=−1.95x

STEP 13

Solve for xxx.x=ln⁡(12)−1.95x = \frac{\ln\left(\frac{1}{2}\right)}{-1.95}x=−1.95ln(21​)​

STEP 14

Calculate the value of xxx.x=ln⁡(0.5)−1.95x = \frac{\ln(0.5)}{-1.95}x=−1.95ln(0.5)​

STEP 15

Use a calculator to find the numerical value of xxx.x=−0.6931−1.95≈0.3554x = \frac{-0.6931}{-1.95} \approx 0.3554x=−1.95−0.6931​≈0.3554

STEP 16

Round the depth to an appropriate number of significant figures, if necessary. Since the problem does not specify, we can leave it as is or round to two decimal places.x≈0.36 metersx \approx 0.36 \text{ meters}x≈0.36 meters

STEP 17

For part (b), we need to find the intensity of light at a depth of 11 meters. Use the equation I=Le−1.95xI = Le^{-1.95x}I=Le−1.95x with x=11x=11x=11.I=Le−1.95⋅11I = Le^{-1.95 \cdot 11}I=Le−1.95⋅11

STEP 18

Calculate the intensity at 11 meters.I=Le−21.45I = Le^{-21.45}I=Le−21.45

STEP 19

For part (c), we need to find the depth at which the intensity is 2.4% of that at the surface. Set I=0.024LI = 0.024LI=0.024L and solve for xxx.0.024L=Le−1.95x0.024L = Le^{-1.95x}0.024L=Le−1.95x

STEP 20

Divide both sides by LLL to isolate the exponential term.0.024=e−1.95x0.024 = e^{-1.95x}0.024=e−1.95x

STEP 21

Take the natural logarithm of both sides to solve for xxx.ln⁡(0.024)=−1.95x\ln(0.024) = -1.95xln(0.024)=−1.95x

STEP 22

Solve for xxx.x=ln⁡(0.024)−1.95x = \frac{\ln(0.024)}{-1.95}x=−1.95ln(0.024)​

STEP 23

Use a calculator to find the numerical value of xxx.x=ln⁡(0.024)−1.95x = \frac{\ln(0.024)}{-1.95}x=−1.95ln(0.024)​

STEP 24

Calculate the depth to an appropriate number of significant figures, if necessary.x≈−3.7337−1.95≈1.9152x \approx \frac{-3.7337}{-1.95} \approx 1.9152x≈−1.95−3.7337​≈1.9152

STEP 25

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(a) Find depth where light intensity is half of surface intensity $L$ . Depth = $D$ meters. (b) Light intensity at 11 m depth = $I = L \cdot e^{(-1.95 \cdot 11)}$ . (c) Find depth where light intensity is 2.4% of surface intensity $L$ . Depth = $D$ meters.

To solve the differential equation, we will use separation of variables. We want to separate the $I$ terms on one side and the $x$ terms on the other side.
$\frac{1}{I}dI = -1.95dx$

Integrate both sides of the equation to find the general solution.
$\int \frac{1}{I}dI = \int -1.95dx$

Perform the integration.
$\ln|I| = -1.95x + C$

Exponentiate both sides to solve for $I$ .
$I = e^{-1.95x + C}$

Since the intensity of light at the surface ( $x=0$ ) is $L$ , we can find the constant $C$ by plugging in $x=0$ and $I=L$ .
$L = e^{C}$

Solve for $C$ .
$C = \ln(L)$

Substitute $C$ back into the equation for $I$ .
$I = e^{-1.95x + \ln(L)}$

Simplify the equation using properties of exponents.
$I = Le^{-1.95x}$

Now we will find the depth at which the intensity is half of $L$ . Set $I = \frac{1}{2}L$ and solve for $x$ .
$\frac{1}{2}L = Le^{-1.95x}$

Divide both sides by $L$ to isolate the exponential term.
$\frac{1}{2} = e^{-1.95x}$

Take the natural logarithm of both sides to solve for $x$ .
$\ln\left(\frac{1}{2}\right) = -1.95x$

Solve for $x$ .
$x = \frac{\ln\left(\frac{1}{2}\right)}{-1.95}$

Calculate the value of $x$ .
$x = \frac{\ln(0.5)}{-1.95}$

Use a calculator to find the numerical value of $x$ .
$x = \frac{-0.6931}{-1.95} \approx 0.3554$

Round the depth to an appropriate number of significant figures, if necessary. Since the problem does not specify, we can leave it as is or round to two decimal places.
$x \approx 0.36 \text{ meters}$

For part (b), we need to find the intensity of light at a depth of 11 meters. Use the equation $I = Le^{-1.95x}$ with $x=11$ .
$I = Le^{-1.95 \cdot 11}$

Calculate the intensity at 11 meters.
$I = Le^{-21.45}$

For part (c), we need to find the depth at which the intensity is 2.4% of that at the surface. Set $I = 0.024L$ and solve for $x$ .
$0.024L = Le^{-1.95x}$

Divide both sides by $L$ to isolate the exponential term.
$0.024 = e^{-1.95x}$

Take the natural logarithm of both sides to solve for $x$ .
$\ln(0.024) = -1.95x$

Solve for $x$ .
$x = \frac{\ln(0.024)}{-1.95}$

Use a calculator to find the numerical value of $x$ .
$x = \frac{\ln(0.024)}{-1.95}$

Calculate the depth to an appropriate number of significant figures, if necessary.
$x \approx \frac{-3.7337}{-1.95} \approx 1.9152$