Find the equation of the line tangent to curve x=sec(t), y=tan(t), at t=pi/6.
I found th slope(derivative of the two which was csc(t). Plugged in Pi/6 and got 2 as the slope.
Now how do I find the equation of the tangent? There are like two equations to work with, I'm confused.
at t=pi/6, y=1/√3 and x=2/√3
So, now you have a point (2/√3,1/√3) and a slope (2), so the line is (using the point-slope form)
y - 1/√3 = 2(x - 2/√3)
To find the equation of the tangent line to the curve x = sec(t), y = tan(t) at t = π/6, you have the slope of the tangent line, which is 2.
Now, you need to find the point on the curve where the tangent line intersects. To do this, substitute t = π/6 into the equations x = sec(t) and y = tan(t):
x = sec(π/6) = 2 √3 / 3
y = tan(π/6) = 1 / √3
So, the coordinates of the point on the curve are (2 √3 / 3, 1 / √3).
Now, you have the slope of the tangent line, m = 2, and a point on the line, (x₁, y₁) = (2 √3 / 3, 1 / √3). To find the equation of the line, you can use the point-slope form:
y - y₁ = m(x - x₁)
Substitute the values:
y - (1 / √3) = 2(x - (2 √3 / 3))
Simplify:
y - (1 / √3) = 2x - 4 √3 / 3
Now, you can rewrite this equation in standard form:
y = 2x - 4 √3 / 3 + 1 / √3
This is the equation of the tangent line to the curve x = sec(t), y = tan(t) at t = π/6.
To find the equation of the tangent line to the curve at t=π/6, you have the slope of the tangent line, which is 2. Now, we need to find the point on the curve that the tangent line passes through at t=π/6.
First, let's find the corresponding x and y coordinates on the curve for t=π/6. Given that x = sec(t) and y = tan(t), we can substitute t=π/6 into these equations:
x = sec(π/6) = 2,
y = tan(π/6) = √3/3.
So, the point on the curve that the tangent line passes through is (2, √3/3).
Now that we have the slope of the tangent line (2) and a point on the line (2, √3/3), we can use the point-slope form of a line equation: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Substituting the values, we get:
y - (√3/3) = 2(x - 2).
Now, we can simplify this equation to obtain the equation of the tangent line in slope-intercept form (y = mx + b) or any other desired form.