math4finance
General
Algebra
Geometry
Coordinate-geometry
Statistics
Calculus
math4finance
math4finance
Home
General
Algebra
Geometry
Coordinate-geometry
Statistics
Calculus
MATH SOLVE
Home
General
Find an equation of the tangent to the curve at the point corresponding to the given value of the pa...
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]