# Help me with differentiation?

A curve has parametric equation: x= 4cos^3 t and y = 4sin^3 t . Find the equation of the normal at t = π/4
asked Jun 17, 2014 in CALCULUS

The parametric equation : x = 4cos3 t and y = 4sin3 t.

When, t = π / 4, x = 4cos3 (π / 4) = 4 * (1/√2)3 = √2

When, t = π / 4, y = 4sin3 (π / 4) = 4 * (1/√2)3 = √2.

Therefore, the point is (√2, √2).

dx / dt = - 12 sin t cos2 t.

dy / dt = 12 sin2 t cos t.

dy / dx = [dy  / dt ] / [dx / dt ] = [12 sin2 t cos t] / [- 12 sin t cos2 t] = - tan t .

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

y ' = - 1.

This is the slope (m ) of the tangent line to the implicit curve at (√2, √2).

The normal line and tangent are perpendecular to each other.

Since the slopes of perpendecular lines are negative reciprocals the slope of nolmal line through the point  (√2, √2) is - 1.

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Slope (m) = 1.

Now, the normal line equation is y = x + b.

Find the y - intercept by substituting the the point in the  line equation say (x, y) = (√2, √2).

√2 = (√2) + b

b = 0.

The normal line equation  is y = x.