Composing Functions and Finding Derivatives

[Lancelot59]

I have two functions:

$$f(x,y,z)=\sqrt{x^{2}+y^{2}+z^{2}}$$
$$\vec{c}(t)=<cos(t),sin(t),1>$$

I need to find:
$$(f \circ c)'(t)$$
and
$$(f \circ c)'(0)$$

I don't have any answers to work with, but I'm guessing I just stick f into c to get this:

$$\vec{c}(t)=<cos(\sqrt{x^{2}+y^{2}+z^{2}}),sin(\sqrt{x^{2}+y^{2}+z^{2}}),1>$$

Then once I have that get the derivative matrix and plug in 0?

 

[lanedance]

what you didn't doesn't make sense

c maps the scalar t to the vector (x(t),y(t),z(t))
f maps the vector (x,y,z) to a scalar f(x,y,z)

so you want to find f(c(t))
f(c(t)) will map t to a scalar

 

[Lancelot59]

what you didn't doesn't make sense

c maps the scalar t to the vector (x(t),y(t),z(t))
f maps the vector (x,y,z) to a scalar f(x,y,z)

so you want to find f(c(t))
f(c(t)) will map t to a scalar

I follow you. So I really should be working with this after I put everything together?

$$f(x,y,z)=\sqrt{(cos(t))^{2}+(sin(t))^{2}+(1)^{2}}$$

 

[lanedance]

looks better, and if you want to include everything explicitly
$$(f \circ c)(t) = f(c(t)) = f(x(t),y(t),z(T))=\sqrt{(cos(t))^{2}+(sin(t))^{2}+(1)^{2}}$$

 

[Char. Limit]

I follow you. So I really should be working with this after I put everything together?

$$f(x,y,z)=\sqrt{(cos(t))^{2}+(sin(t))^{2}+(1)^{2}}$$

Yep. Now just simplify that expression.

 

[Lancelot59]

...I did all that just to get the square root of 2? What a rip off. So the function is just going to be equal to the root of two, and the derivative is zero?

 

[Char. Limit]

...I did all that just to get the square root of 2? What a rip off. So the derivative of the function at any point is therefore a constant root 2?

no, the function at any point is a constant root 2.

 

[HallsofIvy]

$f\circ c(x)$ means f(c(x)) not g(f(x)).