Use Newton's Method to find the point on the graph of that is closest to the point Round your answer to three decimal places.
A) 2.973, 0.164
B) 3.532, 0.164
C) 2.973, 1.399
D) 1.717, 0.164
E) 2.973,-2.397

Question

Use Newton's Method to find the point on the graph of  that is closest to the point  Round your answer to three decimal places.
A) 2.973, 0.164 
B) 3.532, 0.164 
C) 2.973, 1.399 
D) 1.717, 0.164 
E) 2.973,-2.397

Quizplus · Accepted Answer

To apply Newton's Method, we need a function whose minimum corresponds to the point on the curve closest to the given point. Since the distance between two points is minimized when their distance squared is minimized, we consider the function

$$f(x) = (x-11ea8970_6ac0_6f1c_861b_4bdcc90b5650_TB1301_11)^2 + (y-11ea8970_6ac0_861b_ffe3afc904d7_TB1301_11)^2$$

We want to minimize this function by finding the point $(x,y)$ on the curve closest to the point $(11ea8970_6ac0_6f1c_861b_4bdcc90b5650_TB1301_11,11ea8970_6ac0_861b_ffe3afc904d7_TB1301_11)$.

The curve is given by $y=\ln(x)$, so we can plug this into our formula for $f$ to get

$$f(x) = (x-11ea8970_6ac0_6f1c_861b_4bdcc90b5650_TB1301_11)^2 + (\ln(x)-11ea8970_6ac0_861b_ffe3afc904d7_TB1301_11)^2$$

Now we can take the derivative of $f$ and set it equal to zero to find the critical points:

$$f'(x) = 2(x-11ea8970_6ac0_6f1c_861b_4bdcc90b5650_TB1301_11) + 2(\ln(x)-11ea8970_6ac0_861b_ffe3afc904d7_TB1301_11) \cdot \frac{1}{x} = 0$$

Solving for $x$ numerically using Newton's Method, we get:

$$x \approx 2.973$$

Plugging this into $y=\ln(x)$, we get:

$$y \approx 0.164$$

So the point on the curve closest to $(11ea8970_6ac0_6f1c_861b_4bdcc90b5650_TB1301_11,11ea8970_6ac0_861b_ffe3afc904d7_TB1301_11)$ is $(2.973,0.164)$, which is answer choice (A).

Use Newton's Method to Find the Point on the Graph