Angle Between Chord AB and Tangent at Point B on a Curve

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    Calculus Geometry
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Homework Statement



The normal to the curve y= (x + 2)^2 at the point A(-3, 1) meets the curve at B.
Find the angle at B between the curve and chord AB.

2. The attempt at a solution

I found that B is (-0.5, 2.25) and the equation of the normal is 2y= x + 5.
I thought the angle at B between the curve and chord AB was simply the angle between the normal 2y= x + 5 and the gradient function, y'= 2x +4 but the answer is different from what I get.

Any help would be greatful
 
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The angle your are looking is between the chord AB and the tangent of the curve at B.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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