Space curves Find a vector function for the intersection of the given curves. a) z = 4x^2 +y^2 and y = x^2 b) x^2 +y^2 +z^2 = 1 and z^2 = x^2 + y^2
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Space curves
Find a vector function for the intersection of the given curves.
a) z = 4x^2 +y^2 and y = x^2
b) x^2 +y^2 +z^2 = 1 and z^2 = x^2 + y^2
Step by step
Solved in 2 steps with 3 images
- Find the projection of v = <5, 2, -7> onto the line L with vector equation x = td where d = <-5, -3, 6> SEE IMAGEShow clear and detailed solutions • Provide the parametric and symmetric equations of the line that passes through the point (4, -3, 5) and is perpendicular to both vectors (2,6,1) and (1, 3, -5)Find parametric equations for the line that passes through the point (−3, 7) and is perpendicular to the vector <6,−8>.
- Represent the line segment from P(−2, −3, 8), Q(5, 1, −2) by a vector-valued function and by a set of parametric equations.Find parametric and vector equation for line segment which reach (2, 2.4 , 3.5) point and parallel to the 3i + 2j -k vector a) (1+3t)i + (4+2t)j + (3-t)k b) (2+3t)i + (2.4+2t)j + (3.5-t)k c) (1+3t)i + (2+t)j - (3.5-2t)k d) (2t)i + (1.5-2t)j + (3.5-1t)kU'S App -> If the projection of a = î- 2 + 3k on b = 2î+ 1k is zero, then the value of 2 is (A) (B) 1 - 2 (C) 3 - 3 (D) The vector equation of the line passing through the point (-1, 5, 4) and
- A particle moves at constant speed 1445 units along the curve of intersection of the two the two surfaces y = x and z = 2 in the direction of increasing x. Find its velocity when it is at the point (- 12, 144, - 1152). A the point (-12, 144, - 1152), the velocity vector is V = i + + k.5 -Let the initial point of a smooth curve is (2,3, 4) to Terminal point is (6,-5,-4) Find a) The Vector equation of a curve (parametric equation)