Let be the tangent line to the curve y = x^3 at the point (1, 1). Calculate θ correct to the nearest degree.?
also question says...The angle of inclination of is the angle θ that makes with the positive direction of the x-axis. This is a question in my calculus homework. I have looked through my book, but it doesn't say anything about this. If you could explain a bit, it would be greatly appreciated. Any help is appreciated though. Thanks :D
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y = x³
y '(x) = dy/dx = 3x²
y '(1) = 3 (1)² = 3 [ = slope of the tangent at x = 1 ]
Equation of the tangent line to the curve at point (1 ; 1) :
y = 3x + constant
plug the coordinates (1 ; 1) into the equation of the line :
1 = 3 (1) + constant
constant = 1 - 3 = - 2
y = 3x - 2 is the equation of the line.
http://s636.photobucket.com/albums/uu88/stoupils/?...
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Angle θ :
vector (1 ; 0) is one of the vectors on the x-axis
vector (1 ; 3) is one of the vectors on the line y = 3x - 2
Take a look at the graph here :
http://s636.photobucket.com/albums/uu88/stoupils/?...
tan θ = opposite side / adjacent side = 3 / 1 = 3
θ = arctan 3 (or INV TAN on your calculator) =
71.565 degrees or 1.249 radians
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I assume that tangent makes an angle of θ with x - axis.
y = x^3
y ' = 3x^2
slope at x = 1
y '(1) = 3
so tan θ = 3
θ = tan^-1(3)
θ = 71.56 degrees