# Composition Of Functions

• Lancelot59

#### 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?

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

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?

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

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

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.

...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?

...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.

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