**In the exercises 1 - 8, find the linear approximation \(L(x,y)\) and the quadratic approximation \(Q(x,y)\) of each function at the indicated point. These are the \(1^{\text{st}}\)- and \(2^{\text{nd}}\)-degree Taylor Polynomials of these functions at these points. Use a 3D grapher like CalcPlot3D to verify that each linear approximation is tangent to the given surface at the given point and that each quadratic approximation is not only tangent to the surface at the given point, but also shares the same concavity as the surface at this point.**

1) \( f(x,y)=x\sqrt{y},\quad P(1,4)\)

**Answer:**
- \( L(x,y) = 2x+\frac{1}{4}y−1\)

\( Q(x,y) = -1 + 2x + \frac{1}{4}y + \frac{1}{4}(x-1)(y - 4) - \frac{1}{64}(y-4)^2 \)

2) \( f(x,y)=e^x\cos y;\quad P(0,0)\)

3) \( f(x,y)=\arctan(x+2y),\quad P(1,0)\)

**Answer:**
- \( L(x,y) = \frac{1}{4}π−\frac{1}{2} + \frac{1}{2}x+y\)

\( Q(x,y) = \frac{1}{4}π−\frac{3}{4} +x+2y - \frac{x^2}{4} - xy - y^2 \)

4) \( f(x,y)=\sqrt{20−x^2−7y^2},\quad P(2,1)\)

5) \( f(x,y)=x^2y + y^2,\quad P(1,3)\)

**Answer:**
- \( L(x,y) = 12 + 6(x-1) + 7(y - 3) = -15 + 6x + 7y\)

\( Q(x,y) = - 15 + 6x + 7y + 3(x - 1)^2 + 2(x-1)(y - 3) + (y-3)^2 \)

6) \( f(x,y)=\cos x \cos 3y,\quad P(0,0)\)

7) \( f(x,y)=\ln(x^2 + y^2 + 1),\quad P(0,0)\)

**Answer:**
- \( L(x,y) = 0\)

\( Q(x,y) =x^2 + y^2\)

8) \( f(x,y)=\sqrt{2x - y},\quad P(1,-2)\)

9) Verify that the formula for higher-degree Taylor polynomials works for the first-degree Taylor polynomial \(L(x,y) = P_1(x,y)\). For convenience, the formula is given below.

\[P_n(x,y) = \sum_{i=0}^n \sum_{j=0}^{n - i} \frac{\frac{d^{(i+j)}f}{∂x^i∂y^{\,j}}(a,b) }{i!j!}(x-a)^i(y-b)^j\]

10) Determine the new terms that would be added to \(P_3(x,y)\) (which you found in Exercise 13.7.1) to form \(P_4(x,y)\) and determine the fourth-degree Taylor polynomial for one of the functions we've considered and graph it together with the surface plot of the corresponding function in a 3D grapher like CalcPlot3D to verify that it continues to fit the surface better.