Math Snap

STEP 1

What is this asking?
How far does a particle move in 7 seconds if its speed changes over time according to a logarithmic formula?
Watch out!
Velocity can be negative, but distance is always positive!
We need the absolute value of the velocity if we're going backwards.
Luckily, in this problem, the velocity is always positive.

STEP 2

1. Set up the distance integral
2. Solve the integral
3. Calculate the result

STEP 3

Alright, so we're given the velocity function $v(t) = 2 \log_4(t+3)$, and we want to find the total distance traveled between $t=1$ and $t=8$.
Remember, distance is just the integral of the absolute value of velocity.

STEP 4

In our case, $v(t)$ is always positive between $t=1$ and $t=8$, so the absolute value doesn't change anything here.
Phew! That makes our lives easier.
The distance traveled is given by:
$$\int_{1}^{8} |v(t)| dt = \int_{1}^{8} v(t) dt = \int_{1}^{8} 2 \log_4(t+3) dt$$

STEP 5

Let's change the base of the logarithm from base 4 to base e using the change of base formula: $\log_a(x) = \frac{\ln(x)}{\ln(a)}$.
$$\int_{1}^{8} 2 \log_4(t+3) dt = \int_{1}^{8} 2 \cdot \frac{\ln(t+3)}{\ln(4)} dt$$

STEP 6

We can pull the constants out of the integral:
$$\frac{2}{\ln(4)} \int_{1}^{8} \ln(t+3) dt$$

STEP 7

Now, we can use integration by parts to solve the integral of $\ln(t+3)$.
Let $u = \ln(t+3)$ and $dv = dt$.
Then $du = \frac{1}{t+3} dt$ and $v = t$.

STEP 8

The integration by parts formula is $\int u dv = uv - \int v du$, so:
$$\int \ln(t+3) dt = t \ln(t+3) - \int \frac{t}{t+3} dt$$

STEP 9

To solve $\int \frac{t}{t+3} dt$, we can add zero in the numerator: $t = t + 3 - 3$.
$$\int \frac{t}{t+3} dt = \int \frac{t+3-3}{t+3} dt = \int \left( \frac{t+3}{t+3} - \frac{3}{t+3} \right) dt = \int \left( 1 - \frac{3}{t+3} \right) dt$$ $$= t - 3\ln(t+3)$$

STEP 10

Putting it all together:
$$\int \ln(t+3) dt = t \ln(t+3) - (t - 3\ln(t+3)) = t \ln(t+3) - t + 3\ln(t+3)$$

STEP 11

So our definite integral becomes:
$$\frac{2}{\ln(4)} \left[ t\ln(t+3) - t + 3\ln(t+3) \right]_1^8$$

STEP 12

Let's plug in the limits of integration:
$$\frac{2}{\ln(4)} \left[ (8\ln(11) - 8 + 3\ln(11)) - (1\ln(4) - 1 + 3\ln(4)) \right]$$ $$= \frac{2}{\ln(4)} \left[ 11\ln(11) - 8 - 4\ln(4) + 1 \right]$$$$= \frac{2}{\ln(4)} \left[ 11\ln(11) - 4\ln(4) - 7 \right]$$

STEP 13

We can simplify $\ln(4)$ to $2\ln(2)$:
$$\frac{2}{2\ln(2)} \left[ 11\ln(11) - 8\ln(2) - 7 \right]$$ $$= \frac{1}{\ln(2)} \left[ 11\ln(11) - 8\ln(2) - 7 \right]$$

STEP 14

Calculating this gives us approximately:
$$\frac{1}{0.693} [11 \cdot 2.398 - 8 \cdot 0.693 - 7] \approx \frac{1}{0.693} [26.378 - 5.544 - 7] \approx \frac{13.834}{0.693} \approx \textbf{19.96}$$

The distance traveled is approximately 19.96 cm.