Implicit Differentiation

[mace519]

I'm having a lot of trouble with this question. My friend told me one way, but I am not so sure. Here is the question:

http://img483.imageshack.us/my.php?image=calc1bv.jpg
Correct Answer is -3.30units/s

Posible Solution (not correct though):

We need to solve for the roots, and add them so:

. . . . .x = 2(sqrt(c))
. . . . .w = 2(sqrt(c))
. . . . .dw/dt = 2(1/2)(c)^-1/2
. . . . .= 1/sqrt(c) dc/dt
. . . . .= 14/sqrt(14)
. . . . .= +-sqrt(14)
. . . . .= -3.7 units/s

That's my possible solution, but I'm not sure what else I might be able to do. Please advise me. Thank you!

 

[stapel]

Please edit your image. It currently has a very large file size, and trying to view the image directly returns error messages.

Thank you.

Eliz.

 

[mace519]

The motion of a car is enhanced by showing the headlights on a building when the car approaches thebuilding. The headlight 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-axes represent the wall. let the length be measured in metre and time in seconds. Determine the rate at a which the width of the beam is narriwubg when c= -14.

Thats the question. thanks

 

[daon]

The motion of a car is enhanced by showing the headlights on a building when the car approaches thebuilding. The headlight 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-axes represent the wall. let the length be measured in metre and time in seconds. Determine the rate at a which the width of the beam is narriwubg when c= -14.

Thats the question. thanks

This is how I interpret it: The width of the beam will be the distance between the zeros of the quadratic:

\(\displaystyle y = x^2 - 4x + c\)


Using the quadratic equation: y = 0 when x = \(\displaystyle 2 - \sqrt{4-c}\) and \(\displaystyle 2 + \sqrt{4-c}\)

The width is then:

\(\displaystyle w = (2 + \sqrt{4-c}) - (2 - \sqrt{4-c})\)
\(\displaystyle w = 2\sqrt{4-c}\)


So, dw/dt is:

\(\displaystyle \frac{dw}{dt} = \frac{1}{\sqrt{4-c}} (-1)\frac{dc}{dt}\)

Substituting in c = -14 and dc/dt = 14, we get:

\(\displaystyle \frac{dw}{dt} = \frac{1}{\sqrt{4-(-14)}} (-1)(14)\)
\(\displaystyle \frac{dw}{dt} = \frac{-14}{\sqrt{18}}\)

Yeilding \(\displaystyle \frac{dw}{dt} = 2.99831...\)

-daon

 

[mace519]

nvm

 

[stapel]

using the quadractic equation, i keep on getting a different solution for x.

Please reply showing your steps. When you reply, please also provide the answer that you think is correct. (I would assume this is the answer listed in the back of the book. But that answer might be incorrect.)

Thank you.

Eliz.

 

[mace519]

i got it, forgot to factor, btw -14/sqrt(18) = -3.30, which is the correct answer