# Thread: After many hours of trying, I am stuck.

1. ## After many hours of trying, I am stuck.

Theres this one question I was doing for practice that I just cannot do. Even though I know the answer is -0.94, I cant figure out how.

The question goes as followed.

In a computer game, the motion of a car is enhanced by showing headlights on a building, when the car approaches the building. The headlights form a parabola on the road. As the car approaches the building, the beam narrows. The equation of the parabola is y = x^2 - 4x +c, where c < 0 and dc/dt = 14. The x-axis represents the wall, let length be measured in meteres and time in seconds.

Determine the rate at which the width of the beam is narrowing when c = -14.

The answer at the back is -0.94. But I really cant figure out how to get this, and how to approach this question.

help would be appreciated.

2. Originally Posted by Dave J
Theres this one question I was doing for practice that I just cannot do. Even though I know the answer is -0.94, I cant figure out how.

The question goes as followed.

In a computer game, the motion of a car is enhanced by showing headlights on a building, when the car approaches the building. The headlights form a parabola on the road. As the car approaches the building, the beam narrows. The equation of the parabola is y = x^2 - 4x +c, where c < 0 and dc/dt = 14. The x-axis represents the wall, let length be measured in meteres and time in seconds.

Determine the rate at which the width of the beam is narrowing when c = -14.

The answer at the back is -0.94. But I really cant figure out how to get this, and how to approach this question.

help would be appreciated.
This is poorly worded. I will assume "the width of the beam" is actually the
distance between the roots of the quadratic:

$
x^2 - 4x +c
$

The roots of this are:

$
x=\frac{4 \pm \sqrt{16-4c}}{2}
$
,

and the distance between the roots is:

$
\Delta=\sqrt{16-4c}
$

Then:

$
\frac{d \Delta}{dt}=\frac{1}{2}(16-4c)^{-1/2}\ (-4) \frac{dc}{dt}
$

$
\frac{d \Delta}{dt}=-2 \frac{1}{(16-4c)^{1/2}} \frac{dc}{dt}
$

Putting $c=-14$ and $\frac{dc}{dt}=14$, gives:

$
\frac{d \Delta}{dt}=-2 \frac{1}{(16-4c)^{1/2}} \frac{dc}{dt}\approx -3.30
$

which does not agree with your book.

The possibilities are: that I have interpreted the problem incorrectly, I have
made a mistake in my algebra and derivatives, I have made a mistake in the
arithmetic, the book is wrong. No doubt my fellow helpers will comment on
the first three of these.

RonL

3. Hello, Dave!

In a computer game, the motion of a car is enhanced by showing headlights on a building
when the car approaches the building. .The headlights form a parabola on the road.
As the car approaches the building, the beam narrows.
The equation of the parabola is: $y \:= \:x^2 - 4x +c,$ where $c < 0$ and $\frac{dc}{dt} = 14$
The x-axis represents the wall. Length is measured in meters and time in seconds.

Determine the rate at which the width of the beam is narrowing when $c = -14$

I agree completely with Captain Black.
Code:
              |
*      |                        *
|
*     |                       *
---o----+----------------------o---
*  |                    *
c*                 *
|   *         *
|       ***
|

The width of the beam is the distance between the two x-intecepts.

I have the same steps as the Captain . . . and the same answer.
I don't see how we could have misread the problem.
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