Working with Tangent and sin and velocity and curves.

Question

Let ℎ(x)=tan(2x+7). Then

ℎ′(1) is   

and ℎ′′(1) is 

2.Find the derivative of: -3e^-5x cos(-5x).

Now, find the equation of the tangent line to the curve at x=0. Write your answer in mx+b format.

y=

3.A mass attached to a vertical spring has position function given by s(t)=2sin(4t) where t is measured in seconds and s in inches.

Find the velocity at time t=3.  ?  in/s

Find the acceleration at time t=3.  ?  in/s²

Answer 1

1) If y = tan(x) then y ' = sec²(x) so, because you must use the chain rule, h ' (x) = (sec²(2x+7))(2) or 2sec²(2x + 7)

h'(1) = 2sec²(2(1) + 7) = 2sec²(9) = 2[sec(9)]² = 2/[cos(9)]² and, using your calculator (you must be in "radian" mode), we get h'(1) = 2.409

To find h''(1) you must take the second derivative:

2sec²(2x + 7) = 2[sec(2x + 7)]² so h'' = 4[sec(2x + 7)]•(sec(2x+7)tan(2x+7))•(2) = 8sec²(2x + 7)tan(2x + 7)

So h''(1) = 8sec²(2(1) + 7)tan(2(1) + 7) = 8tan(9)/(cos(9))² = -4.359

2) for y = -3e^-5xcos(-5x), we must use the product rule which says (u•v)' = u'v + uv'

In this case u = -3e^-5x so u' = (-3e^-5x)(-5) = 15e^-5x and v = cos(-5x) so v' = (-sin(-5x))(-5) = 5sin(-5x)

So y' = (15e^-5x)(cos(-5x)) + (-3e^-5x)(5sin(-5x)) = 15e^-5xcos(-5x) - 15e^-5xsin(-5x) which you could easily factor into 15e^-5x(cos(-5x) - sin(-5x)).

To find the tangent line at x = 0, first find the y-value when x = 0:

y(0) = -3e^-5(0)cos(-5(0)) = -3(1)•cos(0) = -3 so the tangent point is (0, -3). The slope of the tangent line is the value of the derivative at that point:

y'(0) = 15e^-5(0)(cos(-5(0)) - sin(-5(0))) = 15•1•(cos(0) - sin(0)) = 15•1 = 15

Using the point-slope form of a line, y - y₁ = m(x - x₁) and plugging in x₁ = 0, y₁ = -3, and m = 15 we get: y - -3 = 15(x - 0) or y + 3 = 15x or y = 15x - 3

Given a position function, take the derivative of it to get the velocity function. Then take the derivative a second time to get the acceleration function:

For s(t) = 2sin(4t), v(t) = 8cos(4t) and a(t) = -32sin(4t)

So the velocity at time t = 3 is v(3) = 8cos(4(3)) which you can plug into your calculator (you must be in "radian" mode) to get v(3) = 6.75 in/s

And the acceleration at t = 3 is a(3) = -32sin(4(3)) = 17.17 in²/s