Find the equations of the tangents to the curve x = 9t^2 + 9, y = 6t^3 + 9 that pass through the point (18, 15). y = [ ] (smaller slope) y = [ ] (larger slope)
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This will give us the slope of the tangent to the curve at any point (t). The derivative of y with respect to t is dy/dt = 18t^2. The derivative of x with respect to t is dx/dt = 18t. So, dy/dx = (dy/dt) / (dx/dt) = 18t^2 / 18t = t. Show more…
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