Tangent vector of the curve help

Question

let f(x,y) = 4x^2 - 6x^2 (sqrt(y)) and let γ(t) = <x(t),y(t)> be a curve in the Oxy plane such that at some point t0, we have γ(t0) = (-1,9) and γ'(t0) = (2,-1). Find the tangent vector r'(t0) to the curve r(t) = <x(t),y(t), f(x(t), y(t))> at the point t0. How is this done?

Answer options:

(A). <2, -1, -75>

(B). <-1, 9, -75>

(C). <2, -1, 30>

(D). <-1, -1, 30>

(E). <-36, 3, -1>

Answer 1

here is how

r(t)=(x(t),y(t),z(t)))

r'(t)=(dx/dt,dy/dt,dz/dt)

dz/dt=z_x*(dx/dt)+z_y*(dy/dt)

z_x=8x-12xy^(1/2)

z_y=-3x²y^(-1/2)

it is given at t₀ that x=-1, y=9, dx/dt=2, dy/dt=-1

the answer I compute is r'(t₀)=(2,-1,57)

57=([8(-1)-12(-1)(3)]*2+[-3(1)/3]*(-1)

Can go to 3d curve plotter and use x(t)=2t-3, and y(t)=-t+10.

Let t₀=1. See that the requirements for x,y,x',y' are satisfied

at t=1.

it is given

z(t)=4(2t-3)²-6(2t-3)²(-t+10)^(1/2)

at t=1

r(t)=(-1,9,-14)

v(t)=r'(t)=(2,-1,57)