# Math Help - tangent lines in curves

1. ## tangent lines in curves

1) If , then __16__
Use this to find the equation of the tangent line to the parabola at the point . The equation of this tangent line can be written in the form where __16__
and where _____

but i think b = 27 but apparently its not.

2)For what values of x does the graph of have a horizontal tangent? Enter the x values in order, smallest first, to 4 places of accuracy: ___ ____

I don't really know how to solve this one.

3)
Given that

Calculate

4)A particle moves along a straight line and its position at time is given by where s is measured in feet and t in seconds.
Find the velocity (in ft/sec) of the particle at time : __90__
The particle stops moving (i.e. is in a rest) twice, once when and again when where . is ___
and is ____
What is the position of the particle at time ? __3488___
Finally, what is the TOTAL distance the particle travels between time and time ? _____

5 )If a ball is thrown vertically upward from the roof of foot building with a velocity of ft/sec, then what is the maximum height the ball reaches? ________
What is the velocity of the ball when it hits the ground? ______

Any help is greatly appreciated.

2. for 1)

since $y=2x^2 -4x + 2$
then $\frac{dy}{dx}=4x-4$

For the tangent line at the point (5,34), first we need the slope, give by the derivative, so

$\frac{dy}{dx}=4(5)-4=20-4=16$

Now, for the line equation
$y-35=16(x-5)$
$y-35=16x-80$
$y=16x-80+35=16x-45$

3. for 2) the function will have a horizontal tangent whenever the derivative is zero (makes sense, since at these points, the slope of the tangent is '0')

$f(x)=8x^3-72x^2+216x-96$ so
$f'(x)=24x^2-144x+216=24(x^2-6x+9)=24(x-3)^2$

So it will have a horizontal tangent for x=3.

4. For 3) you need to use the chain rule.
$f(x)=x^5 h(x)$
$f'(x)=x^5 h'(x)+5x^4h(x)$
$f'(-1)=(-1)^5 h'(-1)+5(-1)^4h(-1)$
$f'(-1)=(-1)(6)+5(3)=-6+15=9$

5. 4)A particle moves along a straight line and its position at time is given by where s is measured in feet and t in seconds. ...
The particle stops moving (i.e. is in a rest) twice, once when and again when where . is ___
and is ____
The particle is at rest when the velocity is 0.

Finally, what is the TOTAL distance the particle travels between time and time ? _____
Since velocity is continuous, the particle will change direction only when the velocity of 0. Find the position at each of these times and add up the distances between the positions.

5 )If a ball is thrown vertically upward from the roof of foot building with a velocity of ft/sec, then what is the maximum height the ball reaches? ________
What is the velocity of the ball when it hits the ground? ______
If you know the constant acceleration equations, then use $v^2=u^2+2as$ and remember that at the maximum height the velocity will be 0.

If you don't have this formula, you have:
$\frac{d^2x}{dt^2} = -g$
Integrating both sides with respect to t:
$\frac{dx}{dt} = -gt+c$
at $t= 0, \frac{dx}{dt} = 48$
Substitute this in to find c, then integrate again to get
$x = -gt^2+ct+d$
You can find d using the fact that x=64 when t=0. You can then do the remainder of the question. Remember that only positive values of t are important.