To find the intersection of the two population models, we need to set the two functions equal to each other:
-
The population of Mathville is given by:
\[ m(t) = 100 \cdot \sqrt[3]{t} \] -
The population of Calcfield is given by:
\[ c(t) = 18t \]
To find the intersection, we set these two functions equal to each other:
\[ 100 \cdot \sqrt[3]{t} = 18t \]
Next, to eliminate the cube root, we will first divide both sides by 2 for simplification:
\[ 50 \cdot \sqrt[3]{t} = 9t \]
Next, we can solve for \( t \). We'll isolate \( \sqrt[3]{t} \):
\[ \sqrt[3]{t} = \frac{9t}{50} \]
Now, we can eliminate the cube root by cubing both sides:
\[ t = \left( \frac{9t}{50} \right)^3 \]
Expanding the right side, we calculate:
\[ t = \frac{729t^3}{125000} \]
Next, we multiply both sides by 125000 to eliminate the fraction:
\[ 125000t = 729t^3 \]
Rearranging this equation gives us:
\[ 729t^3 - 125000t = 0 \]
Now, we can factor out \( t \):
\[ t(729t^2 - 125000) = 0 \]
Setting each factor to zero gives:
- \( t = 0 \) (This corresponds to the year 2000.)
- \( 729t^2 - 125000 = 0 \)
Now, let's solve for \( t \) in the second equation:
\[ 729t^2 = 125000 \]
Dividing both sides by 729:
\[ t^2 = \frac{125000}{729} \]
Taking the square root of both sides:
\[ t = \sqrt{\frac{125000}{729}} = \frac{\sqrt{125000}}{\sqrt{729}} = \frac{\sqrt{125 \cdot 1000}}{27} = \frac{10 \sqrt{125}}{27} \]
Calculating \( \sqrt{125} = 5\sqrt{5} \):
\[ t = \frac{10 \cdot 5\sqrt{5}}{27} = \frac{50\sqrt{5}}{27} \]
Approximating \( \sqrt{5} \approx 2.236 \):
\[ t \approx \frac{50 \cdot 2.236}{27} \approx \frac{111.8}{27} \approx 4.14 \]
This indicates the intersection occurs approximately 4.14 years after 2000, or around the year 2004.
Thus, the solution shows that both equations intersect around the year 2004, confirming that the populations of Mathville and Calcfield are approximately equal in that year.