Q. 55 - Calculus

Question

Use the first-order partial derivatives of the functions in Exercises 55–64 to find the equation of the plane tangent to the graph of the function at the indicated point P. Note that these are the same functions as in Exercises 43–52.

$$f(x,y)=x^{2}-y^{2}, P=(1,-3)$$

Step-by-Step Solution

Answer

The equation of the plane tangent to the graph of the given function at the indicated point P is $$2x+6y-z=-8$$.

Step 1. Given Information

$$f(x,y)=x^{2}-y^{2}, P=(1,-3)$$

Step 2. Explanation

Using first-order partial derivatives of the given function with respect to $$x$$, we get

$$f_{x}(1,-3)=2x \mid_{(1,-3)}$$

$$\implies f_{x}(1,-3)=2$$

Using first-order partial derivatives of the given function with respect to $$y$$, we get

$$f_{y}(1,-3)=-2y \mid_{(1,-3)}$$

$$\implies f_{y}(1,-3)=6$$

So, we get $$f(1,-3)=-8$$

Now, the tangent can be given as,

$$f_{x}(1,-3)(x-1)+f_{y}(1,-3)(y+3)=zf(1,-3)$$

Sybstituting the values in the above expression, we get

$$2(x-1)+6(y+3)=z+8$$

$$\implies 2x+6y-z=-8$$

Therefore, the equation of the plane tangent to the graph of the given function at the indicated point P is $$2x+6y-z=-8$$.