Quadratic modelling (problems in context)

Question

1. Question from Photo: http://i.imgur.com/aLe3JU6.png?1 ;

Can you explain how you convert the word problem into algebraic equations.

The Answer is D= 5+0.5(t-2)^2

Answer 1

Hi Davado,

 

The equation has to be of the form y = kx² + 5

 

The x² indicates the parabola and the +5 is the minimum of the parabola.

 

Now, since the minimum is reached after 2 hours, we need to move x to the right by a value of 2.  So our equation becomes

y = k(x-2)² + 5

 

Finally, we need to evaluate k.

The traveler starts 7 km from the road.  This means y = 7 when x = 0

7 = k(0-2)² + 5

We can solve this for k

2 = 4k

1/2 = k

 

So y = 0.5 (x-2)² + 5

 

This is equivalent to the solution you posted.

D= 5+0.5(t-2)^2

 

 

 

Answer 2

Do you know the vertex form of a parabola? It is

 

y = a(x-h)² + k,

 

where (h,k) is the vertex of the parabola.  To adapt this to our problem, we need to identify the dependent and independent variables in the problem, and substitute them in place of y and x in the vertex form above.

 

Our goal is develop a formula which describes distance as a function of time.  Thus, time (t) is the independent variable, and distance (D) is the dependent variable.  So far, we have

 

D = a(t-h)² + k

 

Next, we turn our attention to the graph.  Looking at the curve, we see that the vertical direction represents time (oriented with the positive direction facing up), and the horizontal direction represents distance (oriented with the positive direction facing left).  So the graph setup is a bit different from those you're accustomed to; here, the dependent variable is on the horizontal axis, the positive direction of which faces opposite to its typical representation. 

 

From the graph, we see the vertex is (t,D) = (2,5).  These values can be substituted into the equation:

 

D = a(t-2)² + 5

 

We also have another point which is used to solve for the constant a.  "Initially" means that at time = 0, "her distance from the road is 7 km", or D = 7.  On the graph, this is the point (0,7).  Substituting these values into t and D, respectively, yields

 

7 = a(0-2)² + 5.  Now, we solve for a:

7 = 4a + 5

2 = 4a

.5 = a.

 

Putting this value of a back into the equation, we have

 

D = .5(t-2)² + 5.