# After many hours of trying, I am stuck.

• Dec 3rd 2006, 11:01 AM
Dave J
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.
• Dec 3rd 2006, 11:28 AM
CaptainBlack
Quote:

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:

$\displaystyle x^2 - 4x +c$

The roots of this are:

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

and the distance between the roots is:

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

Then:

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

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

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

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

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
• Dec 3rd 2006, 05:08 PM
Soroban
Hello, Dave!

Quote:

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: $\displaystyle y \:= \:x^2 - 4x +c,$ where $\displaystyle c < 0$ and $\displaystyle \frac{dc}{dt} = 14$
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 $\displaystyle c = -14$

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.