Find the equation of the normal to the curve y= 5tan(x) -3 at the point where x= π/4.
+33661 First get the tangent slope at x = π/4
Slope = dy/dx = 5(tan(x)^2 + 1) tan( π/4) = 1, so slope at x = π/4 is 5*2 = 10
The slope of the normal is -1/the tangent slope, namely -1/10
The equation of the normal is therefore y = -x/10 + c where c is a constant, found from the fact that the line goes through the point ( π/4, 5tan( π/4)-3) or ( π/4, 2)
2 = - π/40 + c so c = 2+ π/40
equation of normal: y = -x/10 + 2+ π/40
+33661 First get the tangent slope at x = π/4
Slope = dy/dx = 5(tan(x)^2 + 1) tan( π/4) = 1, so slope at x = π/4 is 5*2 = 10
The slope of the normal is -1/the tangent slope, namely -1/10
The equation of the normal is therefore y = -x/10 + c where c is a constant, found from the fact that the line goes through the point ( π/4, 5tan( π/4)-3) or ( π/4, 2)
2 = - π/40 + c so c = 2+ π/40
equation of normal: y = -x/10 + 2+ π/40
