- #1
cherryyosh
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Let [itex]F: R^2 \rightarrow R^4[/itex] be
$$F(x,y) = (x^3y,sin(xy),3,xy^3)$$
i) Find the Jacobian matrix of F at (1, pi)
ii) What is the local linear approximation to F at (1, pi)?
iii) Write the equation of the tangent plane to the graph of F(1,pi)
iv)use ii) to compute (0.99, pi+.001) approximately
for i) I got the jacobian being $$\left(\begin{array} 3\pi & 1 \\ -\pi & -1 \\ 0 & 0 \\ \pi^3 & 3\pi^2 \end{array} \right) $$
With a Linear approximation of
$$\left(\begin{array} 3\pi x + y - 3\pi \\ -\pi x - y + 2\pi \\ 3 \\ \pi^3 x + 3\pi^2 y - 3 \pi^3 \end{array} \right) $$
But I am not sure where to go from here to get the tangent plane equation. Any help would be appreciated.
$$F(x,y) = (x^3y,sin(xy),3,xy^3)$$
i) Find the Jacobian matrix of F at (1, pi)
ii) What is the local linear approximation to F at (1, pi)?
iii) Write the equation of the tangent plane to the graph of F(1,pi)
iv)use ii) to compute (0.99, pi+.001) approximately
Homework Equations
The Attempt at a Solution
for i) I got the jacobian being $$\left(\begin{array} 3\pi & 1 \\ -\pi & -1 \\ 0 & 0 \\ \pi^3 & 3\pi^2 \end{array} \right) $$
With a Linear approximation of
$$\left(\begin{array} 3\pi x + y - 3\pi \\ -\pi x - y + 2\pi \\ 3 \\ \pi^3 x + 3\pi^2 y - 3 \pi^3 \end{array} \right) $$
But I am not sure where to go from here to get the tangent plane equation. Any help would be appreciated.