Question

Two videos on youtube were recently uploaded. One is a playful kitten video and the other is a laughing baby video.
The kitten video had
\[10\] views by the end of the first minute after it was uploaded, and the cumulative number of views increases by
\[50\] views by the end of each additional minute.
The baby video had
\[3\] views by the end of the first minute after it was uploaded, and the cumulative number of views increases by a factor of approximately
\[3.5\] by the end of each additional minute.
By the end of which minute will the baby video's cumulative number of views first exceed the kitten video's cumulative number of views?

Answer 1

To analyze and compare the number of views for the kitten video and the laughing baby video over time, let's define the number of views for each video after \( t \) minutes.

### Kitten Video:
- Initial views after 1 minute: \( 10 \)
- Increment after each additional minute: \( 50 \) views per minute

The cumulative number of views for the kitten video after \( t \) minutes can be modeled by the equation:
\[
K(t) = 10 + 50(t - 1) \quad \text{for } t \geq 1
\]
This simplifies to:
\[
K(t) = 50t - 40
\]

### Laughing Baby Video:
- Initial views after 1 minute: \( 3 \)
- Increment after each additional minute: A factor of approximately \( 3.5 \)

The cumulative number of views for the baby video after \( t \) minutes can be modeled by:
\[
B(t) = 3 \cdot (3.5)^{(t-1)} \quad \text{for } t \geq 1
\]

### Finding When Baby Video Exceeds Kitten Video:
We need to find the smallest integer \( t \) such that:
\[
B(t) > K(t)
\]
Which translates to:
\[
3 \cdot (3.5)^{(t-1)} > 50t - 40
\]

We can evaluate this inequality for successive values of \( t \).

1. **For \( t = 1 \)**:
\[
K(1) = 10 \quad \text{and} \quad B(1) = 3 \cdot (3.5)^{0} = 3
\]
\( 3 \not> 10 \)

2. **For \( t = 2 \)**:
\[
K(2) = 50(2) - 40 = 60 \quad \text{and} \quad B(2) = 3 \cdot (3.5)^{1} = 3 \cdot 3.5 = 10.5
\]
\( 10.5 \not> 60 \)

3. **For \( t = 3 \)**:
\[
K(3) = 50(3) - 40 = 110 \quad \text{and} \quad B(3) = 3 \cdot (3.5)^{2} = 3 \cdot 12.25 = 36.75
\]
\( 36.75 \not> 110 \)

4. **For \( t = 4 \)**:
\[
K(4) = 50(4) - 40 = 160 \quad \text{and} \quad B(4) = 3 \cdot (3.5)^{3} = 3 \cdot 42.875 \approx 128.625
\]
\( 128.625 \not> 160 \)

5. **For \( t = 5 \)**:
\[
K(5) = 50(5) - 40 = 210 \quad \text{and} \quad B(5) = 3 \cdot (3.5)^{4} = 3 \cdot 150.0625 \approx 450.1875
\]
\( 450.1875 > 210 \)

At \( t = 5 \), the laughing baby video first exceeds the kitten video. Thus, the answer is:
\[
\boxed{5}
\]