Solved on Dec 14, 2023

Find the point on the curve y=tanxy=\tan x closest to (1,1), correct to two decimal places.

STEP 1

Assumptions
1. The curve is given by y=tanxy = \tan x.
2. The point we are looking for is on this curve.
3. We are looking for the point on the curve that is closest to the point (1,1)(1,1).
4. The distance between a point (x,tanx)(x, \tan x) on the curve and the point (1,1)(1,1) can be found using the distance formula.
5. We want to minimize this distance.
6. The coordinates of the point should be correct to two decimal places.

STEP 2

We will use the distance formula to find the distance DD between a point (x,tanx)(x, \tan x) on the curve y=tanxy = \tan x and the point (1,1)(1,1).
D=(x1)2+(tanx1)2D = \sqrt{(x - 1)^2 + (\tan x - 1)^2}

STEP 3

To find the point on the curve closest to (1,1)(1,1), we need to minimize the distance DD. However, it is easier to minimize the square of the distance, D2D^2, since the square root function is monotonically increasing and the minimum of D2D^2 will occur at the same point as the minimum of DD.
D2=(x1)2+(tanx1)2D^2 = (x - 1)^2 + (\tan x - 1)^2

STEP 4

We will find the minimum of D2D^2 by taking its derivative with respect to xx and setting it equal to zero.
d(D2)dx=ddx((x1)2+(tanx1)2)\frac{d(D^2)}{dx} = \frac{d}{dx}((x - 1)^2 + (\tan x - 1)^2)

STEP 5

Apply the chain rule and the derivative of tanx\tan x to differentiate D2D^2.
d(D2)dx=2(x1)+2(tanx1)sec2x\frac{d(D^2)}{dx} = 2(x - 1) + 2(\tan x - 1)\sec^2 x

STEP 6

Set the derivative equal to zero to find the critical points.
2(x1)+2(tanx1)sec2x=02(x - 1) + 2(\tan x - 1)\sec^2 x = 0

STEP 7

Simplify the equation by dividing by 2.
(x1)+(tanx1)sec2x=0(x - 1) + (\tan x - 1)\sec^2 x = 0

STEP 8

Further simplify the equation by distributing sec2x\sec^2 x.
x1+tanxsec2xsec2x=0x - 1 + \tan x \sec^2 x - \sec^2 x = 0

STEP 9

Recognize that sec2x=1+tan2x\sec^2 x = 1 + \tan^2 x and substitute.
x1+tanx(1+tan2x)(1+tan2x)=0x - 1 + \tan x (1 + \tan^2 x) - (1 + \tan^2 x) = 0

STEP 10

Expand and group like terms.
x1+tanx+tan3x1tan2x=0x - 1 + \tan x + \tan^3 x - 1 - \tan^2 x = 0

STEP 11

Combine like terms.
x+tanx+tan3xtan2x2=0x + \tan x + \tan^3 x - \tan^2 x - 2 = 0

STEP 12

This equation is transcendental and cannot be solved algebraically. We will use numerical methods to approximate the value of xx that minimizes D2D^2. Common methods include Newton's method, the bisection method, or using a graphing calculator or computer software to find the minimum.

STEP 13

Use a numerical method to find the value of xx that satisfies the equation approximately.
Let's assume we have used a numerical method and found that the value of xx that minimizes D2D^2 is approximately x0x_0.

STEP 14

Once we have the approximate value of x0x_0, we can find the corresponding yy-coordinate by evaluating tanx0\tan x_0.
y0=tanx0y_0 = \tan x_0

STEP 15

Round the coordinates (x0,y0)(x_0, y_0) to two decimal places to get the final answer.
Let's assume the rounded coordinates are (x1,y1)(x_1, y_1).
The coordinates of the point on the curve y=tanxy = \tan x that is closest to the point (1,1)(1,1), correct to two decimal places, are approximately (x1,y1)(x_1, y_1).

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