Question

Find f'(2) = u(t) dot v(t), u(2)=<1,2,-1>, u'(2)=<3,0,4>, and v(2)=. Please show work

Answer 1

v(t)=<t,t^(2),t^(3)>

v ' (t) = <1,2t, 3t^2> ... derive each component separately

v ' (2) = <1,2(2), 3(2)^2>

v ' (2) = <1, 4, 12>

Answer 2

now I think you meant to say

f(t) = u(t) + v(t)

and you want to find f ' (2)

Answer 3

if so, then

f(t) = u(t) + v(t)

f ' (t) = u ' (t) + v ' (t)

f ' (2) = u ' (2) + v ' (2)

f ' (2) = <3,0,4> + <1, 4, 12>

I'll let you finish up

Answer 4

I'm sorry, they wanted the dot product.

Answer 5

the dot product between which two vectors

u ' (2) and v ' (2)

?

Answer 6

The question is strange but it wants the dot product of u(t) and v(t), then find f'(2) of that dot product's answer. The book is getting an answer of 35.

Answer 7

f'(2)=u(t) dot v(t)

Answer 8

The v in the question has a 2 in the parenthesis

Answer 9

The mistakes in the question has been corrected.

Answer 10

I tried to find the vector equation that was used to get u(2) and u'(2) but I can't figure it out.

Answer 11

so you're essentially solving for t?

this problem is so bizarre

Answer 12

This is what I got, but I'm not sure on the notation...


u(2)=<1,2,-1>,
u'(2)=<3,0,4>,
v(2)=<t,t^(2),t^(3)>
v'(2) = <1,2t,3t^2>


f(t) = u(t) dot v(t)


f ' (t) = u ' (t) dot v(t) + u(t) dot v ' (t)


f ' (t) = u ' (2) dot v(2) + u(2) dot v ' (2)


f ' (t) = <3,0,4> dot <t,t^(2),t^(3)> + <1,2,-1> dot <1,2t,3t^2>


f ' (t) = 3t+4t^3 + 1+4t-3t^2


f ' (2) = 3(2)+4(2)^3 + 1+4(2)-3(2)^2


f ' (2) = 35