Find the Tangent Given the Point (-1/2,( square root of 3)/2)

Problem

Find the tangent to the unit circle at *(−1/2,√(,3)/2)

Solution

1. Identify the equation of the curve. Since the point (−1/2,√(,3)/2) satisfies x2+y2=1 we are finding the tangent to the unit circle.

2. Differentiate the equation x2+y2=1 implicitly with respect to x to find the slope of the tangent line.

d()/d(x)*(x2+y2)=d()/d(x)*(1)

2*x+2*yd(y)/d(x)=0

3. Solve for the derivative d(y)/d(x) to determine the slope formula.

2*yd(y)/d(x)=−2*x

d(y)/d(x)=−x/y

4. Substitute the coordinates of the given point x=−1/2 and y=√(,3)/2 into the derivative to find the specific slope m

m=−(−1/2)/(√(,3)/2)

m=1/√(,3)

5. Apply the point-slope formula y−(y_1)=m*(x−(x_1)) using the point (−1/2,√(,3)/2) and the slope m=1/√(,3)

y−√(,3)/2=1/√(,3)*(x−(−1/2))

y−√(,3)/2=1/√(,3)*(x+1/2)

6. Simplify the equation into slope-intercept form.

y=1/√(,3)*x+1/(2√(,3))+√(,3)/2

y=√(,3)/3*x+√(,3)/6+(3√(,3))/6

y=√(,3)/3*x+(2√(,3))/3

Final Answer

y=√(,3)/3*x+(2√(,3))/3

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