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find an equation for the heat tangent come curve at the suggest defined by the given value that t 0 votes

x= 3 sin t, y= 3 cos t, t= 3pi/4.

You are watching: Find an equation for the line tangent to the curve at the point defined by the given value of t

The parametric equation is x = 3 sin t, y = 3 cos t, t = 3π / 4.

When, t = 3π / 4, 3 sin (3π / 4) = 3(1 /√2) = 3 / √2.

When, t = 3π / 4, 3 cos (3π / 4) = 3(- 1 /√2) = - 3 / √2.

Therefore, the suggest is (3 / √2, - 3 / √2).

dx / dt = 3 cos t

dy / dt = 3 (- sin t ) = - 3 sin t

dy / dx = <dy / dt > / <dx / dt > = - 3 sin t / 3 cos t = - tan t.

When, t = 3π / 4, dy / dx = - tan (3π / 4) = - (- 1) = 1.

y " = 1.

This is the steep (m ) that the tangent heat to the implicit curve in ~ (3 / √2, - 3 / √2).

To discover the tangent heat equation, instead of the worths of m = 1 and (x, y ) = (3 / √2, - 3 / √2) in the steep intercept kind of an equation.

y = mx + b

- 3 / √2 = 1(3 / √2) + b

b = - 3√2.

Substitute m = 1 and also b = - 3√2 in y = mx + b.

y = 1(x ) - 3√2

The tangent heat equation is y = x - 3√2.