Question

What is the slope of the line normal to the tangent line of `f(x) = sec^2x-xcos(x-pi/4)` at `x= (15pi)/8`?

Answer 1

`=> y = 0.063(x - (15pi)/8) - 1.08`

Interactive graph

Explanation:

The first thing we'll need to do is calculate `f'(x)` at `x = (15pi)/8`.

Let's do this term by term. For the `sec^2(x)` term, note that we have two functions embedded within one another: `x^2`, and `sec(x)`. So, we'll need to use a chain rule here:

`d/dx(sec(x))^2 = 2sec(x) * d/dx(sec(x))`

`color(blue)(= 2sec^2(x)tan(x))`

For the 2nd term, we'll need to use a product rule. So:

`d/dx(xcos(x-pi/4)) = color(red)(d/dx(x))cos(x-pi/4) + color(red)(d/dxcos(x-pi/4))(x)`

`color(blue)(= cos(x-pi/4) - xsin(x-pi/4))`

You may wonder why we didn't use a chain rule for this part, since we have an `(x - pi/4)` inside the cosine. The answer is we implicitly did, but we ignored it. Notice how the derivative of `(x - pi/4)` is simply 1? Hence, multiplying that on doesn't change anything, so we do not write it out in calculations.

Now, we put everything together:

`d/dx(sec^2x-xcos(x-pi/4)) = color(violet)(2sec^2(x)tan(x) - cos(x-pi/4) + xsin(x-pi/4))`

Watch your signs.

Now, we need to find the slope of the line tangent to `f(x)` at `x = (15pi)/8`. To do this, we just plug this value into `f'(x)`:

`f'((15pi)/8) = (2sec^2((15pi)/8)tan((15pi)/8) - cos((15pi)/8-pi/4) + (15pi)/8sin((15pi)/8-pi/4)) = color(violet)(~~-6.79)`

However, what we want is not the line tangent to f(x), but the line normal to it. To get this, we just take the negative reciprocal of the slope above.

`m_(norm) = -1/-15.78 color(violet)(~~0.015)`

Now, we just fit everything into point slope form:

#y = m(x-x_0) + y_0

`=> y = 0.063(x - (15pi)/8) - 1.08`

Take a look at this interactive graph to see what this looks like!

Hope that helped :)