the plane tangent to the surface x^2 + y^2 = 25 at the point (3,4,0)
is what ?
2006-10-30 20:34:10 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
listen at school then you will know
2006-10-30 20:57:26 · answer #1 · answered by Anonymous · 0⤊ 1⤋
i forgot this one but i think we can treat this as a two dimensional problem letting y = +sqrt(25-x^2) --- we take the positive since we are on the first quadrant (3,4)
taking the derivative we get dy/dx = -2x/sqrt(25-x^2) and plugging in x = 3 we get dy/dx = -6/4 = -3/2, which means this is the slope at x = 3.
Next we have to find the y intercept using the equation y = mx+b where m = -3/2, we plug in x = 3 and y = 4 so we get b = 17
so the plane tangent to the surface is y = -3x/2 + 17 (reasonable answer don't you think?)
2006-10-31 04:53:02 · answer #2 · answered by Jeremy 2 · 0⤊ 0⤋
Let's see - the plane that touches the circle at (3,4,0) has a normal going through the origin (0,0,0) and goes through (3,4,0)
Equation of the normal is:
4x-3y = 0
I guess the plane is:
4x - 3y + 0z = 0
2006-10-31 04:55:23 · answer #3 · answered by Orinoco 7 · 0⤊ 0⤋
a mystery to me too.
2006-10-31 04:43:21 · answer #4 · answered by Anonymous · 0⤊ 0⤋