Finding Tangent and Perpendicular Vectors on a 2D Graph

[ZdravkoBG]

Homework Statement

f (x) = e^(3x) + sin(2x) + 3x +1

(a) Find a vector V that is tangent to the graph of y = f(x) at the point ( 0, 2).
(b) Find a vector N that is perpendicular to the graph of y = f(x) at the point ( 0, 2).

2. The attempt at a solution

The first step I took is to find the derivative of the function, since the problem is asking for a tangent at a point.

I got this: f'(x) = 3e^(3x) + 2cos(2x) + 3

However, I am unsure how to continue. The graph of "y = f(x)" is kind of confusing. I am thinking of maybe somehow getting parametric equations for the tangent line, which would allow me to build a vector. But I am unsure how to do this.

 

[ehild]

Hi Zdravko,

You need to use the general equation of a straight line -what is it? ehild

 

[ZdravkoBG]

Well, the general equations would be y=mx+b
However, at this point in the class, we are only using parametrics for lines.

In other words, the only lines we have built so far (involving vectors) have been in this format:

x = a + bt
y = a + bt
etc...

 

[ehild]

Well, the general equations would be y=mx+b
However, at this point in the class, we are only using parametrics for lines.

In other words, the only lines we have built so far (involving vectors) have been in this format:

x = a + bt
y = a + bt
etc...

It is easier to use the first equation. But for both methods, you know one point of the line: (0,2), and need the
the slope m or the tangent vector of the line: For that, evaluate f''(x) = 3e^(3x) + 2cos(2x) + 3 at x=0.

ehild

 

[ZdravkoBG]

I think I got it.

I found the slope of the tangent line at x=0 to be 8.
After that, I constructed a vector from point (0,2) to next point on tangent line (1,10).
That vector is V = <1,8> which satisfies the first part of the problem.

For the second part, I built the equation of the line: y=8x+2

Then I saw in my notes that if put in standard form (-8x + y = 2), the "a" and "b" are the vector perpendicular to the line.

So the normal vector to V would be N = <-8,1> .

Is this correct?

 

[ehild]

It is correct. I just noticed that you needed the vectors instead of equations of lines, tangent or perpendicular to the curve.
You did it right: a tangent vector to f(x) is (v_x,v_y)=(1, f'(x)), and a normal vector is ((n_x,n_y)=(-f'(x), 1).
Anyway, the vectors ( a,b) and (-b, a) are perpendicular to each other. (what is the scalar product of two vectors if they are perpendicular?)

ehild

 

[ZdravkoBG]

I forgot about that method to check. The scalar/dot product of the two perpendicular vectors should be 0. Thanks for all the help!

 

[ehild]

I forgot about that method to check. The scalar/dot product of the two perpendicular vectors should be 0. Thanks for all the help!

You are welcome

ehild