Direction

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Oct 27th 2009, 01:32 AM
thekiterunner

Direction

The planned path of a UFO is defined by the equation

y=0.25x^4 + 0.6667x^3 for x > 0

units are in km

(a) Find the direction of motion when the x-value is 2 and 3

(b) Find a point on the UFOs path where it is inclined at 45 degrees to the positive x-axis.

(c) Are there any other points that are similar to (b).

If someone could help i would be very greatful
Oct 27th 2009, 04:47 AM
Open that Hampster!

You're given a distance formula.

The derivative of distance is velocity, so for (a) derive the equation and find the sign of the velocity equation at those points.

For part b/c, just picture a 45-45-90 triangle, just for reference. Each leg of the triangle has a length of $\frac{\sqrt{2}}{2}*x$ , but in this calculation x doesn't matter.

So if you draw a triangle which has a left endpoint connected to the origin, then the hypotenuse becomes a line with a slope 45 degrees with respect to the x axis.

$\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$

So for (b) you just need to find where on x > 0 the slope of y = 1. Which should be easy assuming you have a basic understanding of these concepts.
Oct 27th 2009, 04:51 AM
HallsofIvy

Quote:

Originally Posted by thekiterunner

The planned path of a UFO is defined by the equation

y=0.25x^4 + 0.6667x^3 for x > 0

units are in km

(a) Find the direction of motion when the x-value is 2 and 3

Is this one or two questions. If you mean "Find the directions of motion when x= 2 and x= 3" then the angle the direction makes with the x-axis is the arctangent of the derivative. And with $y= \frac{1}{4}x^4+ \frac{2}{3}x^3$ , the derivative is $y'= x^3+ 2x^2$ . At 2 and 3 the derivatives are 16 and 45 respectively.

[quote](b) Find a point on the UFOs path where it is inclined at 45 degrees to the positive x-axis.[quote]
tan(45)= 1 so you want to solve $y'= x^3+ 2x^2= 1$ .

Quote:

(c) Are there any other points that are similar to (b).

$y'= x^3+ 2x^2= 1$ , being cubic, may have three solutions. I guess that is what is meant by "similar to (b)".

Quote:

If someone could help i would be very greatful
Oct 27th 2009, 01:18 PM
thekiterunner

So that means solving for x^3 +2x^2 = 1

is X^3 + 2x^2 -1
(x+1)(x^2+x-1)

so x = -1
x=0.618
x=-1.618

Another way of say it is
x^3 +2x^2=x

All the points where X^3 +2x^2 crosses the line function x

then I get

x = 0.414
x=-2.414
x=0

Which one of the answers is correct?