Math / Calculus

QuestionA sample of 9 grams of radioactive material is placed in a vault. Let $P(t)$ be the amount remaining after $t$ years, and let $P(t)$ satisfy the differential equation $P^{\prime}(t)=-0.032 P(t)$ . Answer parts $(a)$ through $(g)$ . (a) Find the formula for $P(t)$ . $P(t)=\square$ (Type an expression using t as the variable.)

Studdy Solution

STEP 1

What is this asking? We've got some radioactive material decaying, and we need to figure out how much is left after a certain amount of time. Watch out! Don't mix up the decay constant with the actual amount of material!

STEP 2

1. Set up the differential equation.
2. Solve for $P(t)$ .
3. Determine the constant using the initial condition.

STEP 3

We're given the differential equation $P'(t) = -0.032 \cdot P(t)$ .
This tells us how the amount of material, $P(t)$ , changes over time.
The negative sign indicates it's decaying.

STEP 4

This differential equation screams "exponential decay"!
We can rewrite it as $\frac{P'(t)}{P(t)} = -0.032.$

STEP 5

Now, we can integrate both sides with respect to $t$ : $\int \frac{P'(t)}{P(t)} \, dt = \int -0.032 \, dt.$

STEP 6

The left side integrates to $\ln|P(t)|$ , and the right side to $-0.032 \cdot t + C$ , where $C$ is the constant of integration.
So we have $\ln|P(t)| = -0.032 \cdot t + C.$

STEP 7

To get rid of the natural log, we can exponentiate both sides: $P(t) = e^{-0.032 \cdot t + C}.$

STEP 8

Using properties of exponents, we can rewrite this as $P(t) = e^C \cdot e^{-0.032 \cdot t}.$ Since $e^C$ is just another constant, let's call it $A$ .
This gives us a cleaner formula: $P(t) = A \cdot e^{-0.032 \cdot t}.$

STEP 9

We know that at the beginning ( $t = 0$ ), we have 9 grams of material.
So, $P(0) = 9$ .
Let's plug this into our formula: $9 = A \cdot e^{-0.032 \cdot 0}.$

STEP 10

Since $e^0 = 1$ , we have $9 = A \cdot 1$ , which means $A = 9$ !

STEP 11

Now we can plug this back into our formula for $P(t)$ : $P(t) = 9 \cdot e^{-0.032 \cdot t}.$

STEP 12

The formula for $P(t)$ is $P(t) = 9e^{-0.032t}$ .

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Studdy Solution

STEP 1

What is this asking? We've got some radioactive material decaying, and we need to figure out how much is left after a certain amount of time. Watch out! Don't mix up the decay constant with the actual amount of material!

STEP 2

1. Set up the differential equation.
2. Solve for $P(t)$ .
3. Determine the constant using the initial condition.

STEP 3

We're given the differential equation $P'(t) = -0.032 \cdot P(t)$ .
This tells us how the amount of material, $P(t)$ , changes over time.
The negative sign indicates it's decaying.

STEP 4

This differential equation screams "exponential decay"!
We can rewrite it as $\frac{P'(t)}{P(t)} = -0.032.$

STEP 5

Now, we can integrate both sides with respect to $t$ : $\int \frac{P'(t)}{P(t)} \, dt = \int -0.032 \, dt.$

STEP 6

The left side integrates to $\ln|P(t)|$ , and the right side to $-0.032 \cdot t + C$ , where $C$ is the constant of integration.
So we have $\ln|P(t)| = -0.032 \cdot t + C.$

STEP 7

To get rid of the natural log, we can exponentiate both sides: $P(t) = e^{-0.032 \cdot t + C}.$

STEP 8

Using properties of exponents, we can rewrite this as $P(t) = e^C \cdot e^{-0.032 \cdot t}.$ Since $e^C$ is just another constant, let's call it $A$ .
This gives us a cleaner formula: $P(t) = A \cdot e^{-0.032 \cdot t}.$

STEP 9

We know that at the beginning ( $t = 0$ ), we have 9 grams of material.
So, $P(0) = 9$ .
Let's plug this into our formula: $9 = A \cdot e^{-0.032 \cdot 0}.$

STEP 10

Since $e^0 = 1$ , we have $9 = A \cdot 1$ , which means $A = 9$ !

STEP 11

Now we can plug this back into our formula for $P(t)$ : $P(t) = 9 \cdot e^{-0.032 \cdot t}.$

STEP 12

The formula for $P(t)$ is $P(t) = 9e^{-0.032t}$ .

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Studdy Solution

STEP 1

What is this asking? We've got some radioactive material decaying, and we need to figure out how much is left after a certain amount of time. Watch out! Don't mix up the decay constant with the actual amount of material!

STEP 2

1. Set up the differential equation.2. Solve for P(t)P(t)P(t).3. Determine the constant using the initial condition.

STEP 3

We're given the differential equation P′(t)=−0.032⋅P(t)P'(t) = -0.032 \cdot P(t)P′(t)=−0.032⋅P(t).This tells us how the amount of material, P(t)P(t)P(t), changes over time.The negative sign indicates it's *decaying*.

STEP 4

This differential equation screams "exponential decay"!We can rewrite it as P′(t)P(t)=−0.032.\frac{P'(t)}{P(t)} = -0.032.P(t)P′(t)​=−0.032.

STEP 5

Now, we can **integrate both sides** with respect to ttt: ∫P′(t)P(t) dt=∫−0.032 dt.\int \frac{P'(t)}{P(t)} \, dt = \int -0.032 \, dt.∫P(t)P′(t)​dt=∫−0.032dt.

STEP 6

The left side integrates to ln⁡∣P(t)∣\ln|P(t)|ln∣P(t)∣, and the right side to −0.032⋅t+C-0.032 \cdot t + C−0.032⋅t+C, where CCC is the constant of integration.So we have ln⁡∣P(t)∣=−0.032⋅t+C.\ln|P(t)| = -0.032 \cdot t + C.ln∣P(t)∣=−0.032⋅t+C.

STEP 7

To get rid of the natural log, we can **exponentiate both sides**: P(t)=e−0.032⋅t+C.P(t) = e^{-0.032 \cdot t + C}.P(t)=e−0.032⋅t+C.

STEP 8

STEP 9

We know that at the beginning (t=0t = 0t=0), we have **9 grams** of material.So, P(0)=9P(0) = 9P(0)=9.Let's plug this into our formula: 9=A⋅e−0.032⋅0.9 = A \cdot e^{-0.032 \cdot 0}.9=A⋅e−0.032⋅0.

STEP 10

Since e0=1e^0 = 1e0=1, we have 9=A⋅19 = A \cdot 19=A⋅1, which means A=9A = 9A=9!

STEP 11

Now we can plug this back into our formula for P(t)P(t)P(t): P(t)=9⋅e−0.032⋅t.P(t) = 9 \cdot e^{-0.032 \cdot t}.P(t)=9⋅e−0.032⋅t.

STEP 12

The formula for P(t)P(t)P(t) is P(t)=9e−0.032tP(t) = 9e^{-0.032t}P(t)=9e−0.032t.

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1. Set up the differential equation.
2. Solve for $P(t)$ .
3. Determine the constant using the initial condition.

We're given the differential equation $P'(t) = -0.032 \cdot P(t)$ .
This tells us how the amount of material, $P(t)$ , changes over time.
The negative sign indicates it's decaying.

This differential equation screams "exponential decay"!
We can rewrite it as $\frac{P'(t)}{P(t)} = -0.032.$

Now, we can integrate both sides with respect to $t$ : $\int \frac{P'(t)}{P(t)} \, dt = \int -0.032 \, dt.$

The left side integrates to $\ln|P(t)|$ , and the right side to $-0.032 \cdot t + C$ , where $C$ is the constant of integration.
So we have $\ln|P(t)| = -0.032 \cdot t + C.$

To get rid of the natural log, we can exponentiate both sides: $P(t) = e^{-0.032 \cdot t + C}.$

We know that at the beginning ( $t = 0$ ), we have 9 grams of material.
So, $P(0) = 9$ .
Let's plug this into our formula: $9 = A \cdot e^{-0.032 \cdot 0}.$

Since $e^0 = 1$ , we have $9 = A \cdot 1$ , which means $A = 9$ !

Now we can plug this back into our formula for $P(t)$ : $P(t) = 9 \cdot e^{-0.032 \cdot t}.$

The formula for $P(t)$ is $P(t) = 9e^{-0.032t}$ .