You're on the right track
If you extend the height and slant height, we have a "missing" cone. Let's call the missing cone's slant height x.
So now we can think of a proportion: smaller missing cone to bigger overall cone (actual + missing)
small radius/big radius...
Right and left behavior refer to what graph approaches as x goes to infinity (right) and negative infinity (left)
if the function is 5 minus 7/(2x-3x^2), then when x gets big positive or big negative 7/(2x-3x^2) goes to 0
and we are just left with 5. So the graph approaches y = 5 (as...
when you have a limit where the base and exponent both have a variable, we need to use a log, then L'hospitals Rule
so first let's do ln of our limit to get x out of the exponent
ln [lim x->oo (x/1+x)^x] = x * ln[lim x->oo x/1+x] = lim x->oo[x*ln(x/1+x)]
To find the normal vector to a plane, all we have to do is look at the coefficients
So the normal vector is 3i + 2j + 6k
Speed is the magnitude of the velocity function
so first we need to find v(t) = r ' (t)
r ' (t) = 3e^3t i - 3e^-3t j + (3te^3t + e^3t)k
so v(0) = r ' (0) = 3e^0 i - 3^0 j + (3*0e^0 + 3^0)k = 3i - 3j + k
so then magnitude of velocity at t = 0 is sqrt(3^2...
so if acceleration is a(t) = ti + j + tk
velocity is the antiderivative of acceleration, so we just do antiderivative of each term of acceleration
a(t) = ti+j+tk
v(t) = (1/2t^2+C1)i + (t+C2)j + (1/2t^2+C3)k, where C1, C2, and C3 are the three components...
We can factor 1+6t^2+9t^4 as (1+3t^2)^2
So then we have sqrt [ (1+3t^2)^2) ]
which is just 1+3t^2, which we can now integrate from -1 to 1