This has already been put up but I recieved a comment stating that there is info

Question

This has already been put up but I recieved a comment stating that there is information missing and the thread was closed without an answer. This is the exact question from the text book 'A First Course in Calculus' Chapter 18.2 Q13 if I remember correctly and this is all the information that my teacher has given me.

a) A differentiable curve r(t) lies on the surface
x2+4y2+9z2=14,

and is so parameterized that r(0)=(1,1,1). Let

f(x,y,z)=x2+4y2+9z2

and let h(t)=f(r(t)). Find h'(0).

b) Let g(x,y,z)=x2+y2+z2 and let k(t)=g(r(t)). Suppose in addition that r'(0)=(4,-1,0). Find k'(0).

Explanation / Answer

If r(t) = <x(t) , y(t) , z(t)>

Then since x^2 + 4y^2 + 9z^2 is a constant value of 14

So, f(r(t)) will also be equal to 14

So, h(t) = 14

And thus, h'(t) = 0

So, h'(0) = 0 ----> ANSWER to a

-------------------------------------------------------------------------------------------

g(x,y,z) = x^2 + y^2 + z^2

del g = <gx , gy , gz>

gx = 2x,
gy = 2y and gz = 2z

So, del g = <2x , 2y , 2z>

r'(0) = (4 , -1 , 0)

Integrating :

r(0) = (4t , -t , 0)

x = 4t , y = -t , z = 0

g(t) = 16t^2 + t^2 + 0^2   = 17t^2

g'(t) = 34t

g'(0) = 0

k(t) = g(r(t))

Deriving :

k'(t) = g'(r(t)) * r'(t)

k'(0) = g'(r(0)) * r'(0)

k'(0) = 0 * r'(0)

k'(0) = 0 ----> ANSWER

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