Q. 33 - Calculus

Question

In Exercises 29–34, find the equation of the line tangent to the surface at the given point P and in the direction of the given unit vector u. Note that these are the same functions, points, and vectors as in Exercises 21–26.

$$f(x,y)=\sqrt \frac{y}{x}$$ at $$P=(4,9), u=(-\frac{\sqrt17}{17}, -\frac{4 \sqrt 17}{17})$$

Step-by-Step Solution

Answer

The values of x, y, and z are $$4-\frac{\sqrt 17}{17}t$$, $$9-\frac{4\sqrt 17}{17}t$$, $$\frac{3}{2}-\frac{7\sqrt 17}{816}t$$ respectively.

Step 1. Given Information

$$f(x,y)=\sqrt \frac{y}{x}$$ at $$P=(4,9), u=(-\frac{\sqrt17}{17}, -\frac{4 \sqrt 17}{17})$$

Step 2. Explanation

The equation of the tangent line is given as, $$f_{x}(4,9)(x-4)+f_{y}(4,9)(y-9)=z-f(4,9)$$

Substituting the values, we get

$$-9x+4y-48z=-76$$

The directional derivative of the normal line equation can be given as,

$$\bigtriangledown f(4,9)\cdot u=(\frac{df}{dx}\mid _{(4,9)}i+\frac{df}{dy}\mid _{(4,9)}j)\cdot (-\frac{\sqrt17}{17}i--\frac{4\sqrt17}{17}j)$$

Solving further, we get

$$\bigtriangledown f(4,9)\cdot u=-\frac{7}{816}\sqrt 17$$

Hence, we get the values of x, y, and z as,

$$x=4-\frac{\sqrt 17}{17}t$$

$$y=9-\frac{4\sqrt 17}{17}t$$

$$z=\frac{3}{2}-\frac{7\sqrt 17}{816}t $$