Suppose the vector-valued function r(t) = < f(t) ,g(t) ,h (t) > is smooth on an interval containing the point < f(t0) ,g(t0) ,h(t0) > . The line tangent to r(t) at t = t0 is the line parallel to the tangent vector r'(t0) that passes through < f(t0),g(t0),h(t0)> . Find the line tangent to the curve at t = t0. r(t)=(2t-1,9t+5,t^2), t0= 1 The tangent line is < , , > .

Step-by-step explanation:

r' = <f',g',h'> = <2,9,2t>

So, with t0=1,

r(1) = <1,14,1>

r'(1) = <2,9,2>

Now, what is the line through r(1), in the direction r'(1) ?

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

the correct answer is a.) bc< mn 5 stars if this you : )

step-by-step explanation: