How Do You Find Unit Vectors Parallel and Normal to a Curve at a Specific Point?

fsm
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Homework Statement


Find a unit vector that is (a) parallel to and (b) normal to the graph of f(x) at the given point. Then sketch.

f(x)=x^2
point=(3, 9)

Homework Equations


None that I'm aware of.

The Attempt at a Solution


Find parallel or perpendicular lines, planes, vectors, etc. to a given function has always been a problem for me. I never know where to start. Is it a matter of slope? If so, then how?
 
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Okay, let's start with (a); first we must find an equation for a line parallel to the function at this point, that is the equation of the tangent at this point. How do you suppose we can do this?
 
Take the derivative of the function at 2?
 
fsm said:
Take the derivative of the function at 2?
Close, but why at x=2?
 
fsm said:

Homework Statement


Find a unit vector that is (a) parallel to and (b) normal to the graph of f(x) at the given point. Then sketch.

f(x)=x^2
point=(3, 9)

Homework Equations


None that I'm aware of.
How about f '(x0) is the slope of the tangent line to y= f(x) at xb0, two lines are parallel if they have the same slope, and two lines are normal if the product of their slopes is -1?


The Attempt at a Solution


Find parallel or perpendicular lines, planes, vectors, etc. to a given function has always been a problem for me. I never know where to start. Is it a matter of slope? If so, then how?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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