1. Find the points P and Q on the parabola y=1-x^2 so that the triangle ABC formed by the x-axis and the tangent lines at P and Q is an equilateral triangle.
2. Find the point where the curves y=x^3-3x+4 and y=3(x^2-x) are tangent to each other, that is, have a common tangent line.
3.A car is traveling at night along a highway shaped like a parabola with its vertex at the origin. The car starts at a point 100 m west and 100 m north of the origin and travels in an easterly direction. There is a statue located 100 m east and 50 m north of the origin. At what point on the highway will the car's headlights illuminate the statue?
4. Use linear approximation (or differentials) to extimate the given number:
sin(1 degree)
Answers to all or any of the questions would be great! Thanks!!!!
2007-10-17 05:16:27 · 2 answers · asked by jellybean0424 2 in Science & Mathematics ➔ Mathematics
These are four questions. I will answer one.
1)
Such an equilateral triangle will have sides with slopes ±√3
A line tangent to the parabola at point (x,y) has slope
dy/dx = -2x
So, -2x = ±√3
x = ±√3/2
and for each y = 1-x^2 = 1/4
The points are (√3/2, 1/4) and (-√3/2, 1/4)
2007-10-17 05:57:23 · answer #1 · answered by Scott R 6 · 1⤊ 0⤋
2) intercept
x^3-3x+4=3x^2-3x
x^3-3x^2+4=0 so (x+1)(x-2)^2=0
As x=2 is a double rout at the point (2,6) they are tangent
Verify y´(2)= 3(2)^2-3 = 3(2*2-1)=9
4)The Mc Laurin formula applies with angles expressed in radians
1º=pi/180 rad
a linear approximation for small angles is sin x= x
so sin 1=0.0175
You can do Nº3
2007-10-17 06:45:08 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋