Question

Question Determine the equation of the tangent line at t = 1 to the curve defined parametrically by the equations x(t)=-6√5t and y(t)=-5√√3t+3, for t≥ 0. Enter your answer as an exact answer.

Answer 1

To find the equation of the tangent line at t = 1, we need to find the slope of the curve at that point and use it in the point-slope form. The equation of the tangent line at t = 1 is y - (-5√√3) = -6√5(x - (-6√5)).

Step-by-step explanation:

To find the equation of the tangent line at t = 1, we need to find the slope of the curve at that point. First, let's find the derivatives of x(t) and y(t) with respect to t:

x'(t) = -6√5

y'(t) = -5√√3

Now we can find the slope at t = 1 by substituting it into the derivatives:

x'(1) = -6√5

y'(1) = -5√√3

So, the slope of the curve at t = 1 is -6√5 for x(t) and -5√√3 for y(t). The equation of a line with slope m passing through the point (x₀, y₀) is given by y - y₀ = m(x - x₀). Substituting the values, we get the equation of the tangent line as:

y - (-5√√3) = -6√5(x - (-6√5))

Answer 2

To find the equation of the tangent line at t = 1, we need to find the values of x and y at t = 1. Then, we can calculate the slope of the tangent line and substitute it along with the values of x and y into the point-slope form of a line equation.

Step-by-step explanation:

To find the equation of the tangent line at t = 1 to the parametric curve defined by x(t) and y(t), we follow these steps:

Find the coordinates of the point of intersection (t = 1):x(1) = -6√5 × 1 = -6√5

y(1) = -5√√3 × 1 + 3 = -5√3 + 3

Therefore, the point of intersection is (-6√5, -5√3 + 3).

Find the slope of the tangent line:

The slope of the tangent line at a point can be found using the following formula:

m = dy/dx = (dy/dt) / (dx/dt)

m = (-5/√3) / (-6√5) = √3 / 6

Form the equation of the tangent line:

The general equation of a straight line passing through the point (a, b) with slope m is:

y - b = m(x - a)

Plugging in the values (-6√5, -5√3 + 3) for (a, b) and √3 / 6 for m, we get:

y - (-5√3 + 3) = (√3 / 6)(x - (-6√5))

Simplifying the equation:

y + 5√3 - 3 = (√3 / 6)x + 5√5

Therefore, the equation of the tangent line at t = 1 to the given curve is: y + 5√3 - 3 = (√3 / 6)x + 5√5.