MATH SOLVE

4 months ago

Q:
# Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = t3 + 1, y = t10 + t; t = β1

Accepted Solution

A:

The equation of the tangent line is:

[tex]y - y_t = \frac{y'(t)}{x'(t)}(x - x_t)[/tex]

differentiate x and y

[tex]y - y_t = \frac{10t^9 +1}{3t^2} (x - x_t)[/tex]

Sub in t = -1

[tex]y -0 = \frac{-9}{3}(x - 0) \\ \\ y = -3x[/tex]

[tex]y - y_t = \frac{y'(t)}{x'(t)}(x - x_t)[/tex]

differentiate x and y

[tex]y - y_t = \frac{10t^9 +1}{3t^2} (x - x_t)[/tex]

Sub in t = -1

[tex]y -0 = \frac{-9}{3}(x - 0) \\ \\ y = -3x[/tex]