Hi there, this is calc 2.

See question in attached photo. I am beyond confused on the format of this question. What does it mean by "line tangent to the curve of intersection..."?

I am a visual learner. Can you please sketch a picture of what the question is asking? For ex. can you please sketch a picture of the "line tangent of the curve of intersection of the surface, z and the indicated plane(s) at (-2,4,8)?

If you choose to solve this, can you please explain all your steps? Thanks for your help.

Find the slope of the line tangent to the curve of intersection of the surface z = J4x? + 3y? and the indicated plane at the point (-2.4,8) on the surface :. (@) y=4 (b)x =-2 Given the surface z = \\4x2 + 3y2, we can find the partial derivatives: az 1 4x 8x = 21 4x2 + 3y2 V4x2 + 3y2 az 1 3y 6y = ay 2\\4x2 + 3y2 V4x2 + 3y2 So, the gradient vector Vz is: 4x 3y Vz = 4x2 + 3y2' V4x2 + 3y2 (a) Slope when y = 4 For the plane y = 4, the curve of intersection is given by fixing y = 4, and z = V4x2 + 48. The slope of the tangent line in the x-direction is given by the derivative of z with respect to x at (-2, 4): dz 4x dx V4x2 + 48 Substitute * = -2 into the equation: dz 4(-2) 8 8 8 -1 dx V4(-2)2 + 48 V16 + 48 V64 8 So, the slope of the tangent line when y = 4 is -1(b) Slope when z = 2 For the plane x = 2, the curve of intersection is given by fixing x = 2, and The slope of the tangent line in the y-direction is given by the derivative of z with respect to y at (2, 4): 0P AP RYT T TN Ty Substitute y = 4 into the equation: EA 3(4) N 12 B dy /16+3(4)2 +16+48 / So, the slope of the tangent line when z = 2is 5