Find a vector equation of the line tangent to the graph of r(t) at the point Po on the curve.

- r(t)= 4costi - 3tj ; Po (2, -pi)

- r(t)= t^2i-(1/1+t)j+(4-t^2)k ; Po= (4,1,0)

Question

Find a vector equation of the line tangent to the graph of r(t) at the point Po on the curve.

-> r(t)= 4costi - 3tj ; Po (2, -pi)

-> r(t)= t^2i-(1/1+t)j+(4-t^2)k ; Po= (4,1,0)

anonymous · Answer

@eseidl

turingtest · Answer

take the derivative of the vector function $$\frac d{dt}\vec r(t)$$

anonymous · Answer

that i did...what value do i put in ? 2 or -pi...? @TuringTest

turingtest · Answer

find the value of t where those points occur from r(t) first

turingtest · Answer

once you know the value of t then plug that value back into r(t)

turingtest · Answer

back into r'(t)

anonymous · Answer

The x component of the point (i.e., 2) is your $$\hat{i}$$component.  So,$$4\cos t=2$$in the first equation.  Similarly, -pi is the j component.

anonymous · Answer

you better get the same value of t, using both:$$4\cos t=2$$and/or,$$-3t=-\pi$$

anonymous · Answer

got this part !..what about the second ? 

t^2=4 
t= +- 2

do i put the positive value in the differentiated equation or the negative one ? @eseidl

turingtest · Answer

look at your second component to answer that

anonymous · Answer

Well, you can't tell just looking at the i component.

anonymous · Answer

@TuringTest correct.

anonymous · Answer

Which t values is consistent with$$\frac{-1}{1+t}=1$$?

anonymous · Answer

gotto go.  You should have enough to answer this now.