# Thread: Derivative and part of word problems...

1. ## Derivative and part of word problems...

1. Given that

Calculate 6
Calculate -8
Calculate ?
Calculate ?
All the answers i got for (f*g) and (f/g) are wrong??

2. A space traveler is moving from left to right along the curve
When she shuts off the engines, she will continue traveling along the tangent line at the point where she is at that time. At what point
(x,y) should she shut off the engines in order to reach the point ?
If she was traveling from right to left she would have to shut off the engines at the point (x,y)?

3. If an arrow is shot straight upward on the moon with a velocity of 78 m/s, its height (in meters) after t seconds is given by .
What is the velocity of the arrow (in m/s) after 7 seconds? 66.38 m/s
After how many seconds will the arrow hit the moon? 94 seconds
With what velocity (in m/s) will the arrow hit the moon? ?? (How do i calculate for this?)

2. Originally Posted by Kayla_N
...

2. A space traveler is moving from left to right along the curve
When she shuts off the engines, she will continue traveling along the tangent line at the point where she is at that time. At what point
(x,y) should she shut off the engines in order to reach the point ?
If she was traveling from right to left she would have to shut off the engines at the point (x,y)?

...
Let $\displaystyle T(t, t^2)$ denote the point where the engine is shut off. The slope of the tangent in T is $\displaystyle m_t = 2t$

Then the tangent at the curve in T is:

$\displaystyle y - t^2 = 2t(x-t) ~\implies~ y = 2tx-t^2$

This tangent passes throught the point P(4, 15), that means it's coordinates must satisfy the equation of the tangent:

$\displaystyle 15 = 2t \cdot 4 - t^2~\implies~t^2-8t+15=0~\implies~\boxed{t = 3~\vee~t = 5}$

Therefore the two tangents are:

$\displaystyle t_1:y = 6x-9$ or $\displaystyle t_2: y = 10x-25$

and the tangent points are:

$\displaystyle T_1(3, 9)$ or $\displaystyle T_2(5, 25)$

3. Originally Posted by Kayla_N
...

3. If an arrow is shot straight upward on the moon with a velocity of 78 m/s, its height (in meters) after t seconds is given by .
What is the velocity of the arrow (in m/s) after 7 seconds? 66.38 m/s ....... OK
After how many seconds will the arrow hit the moon? 94 seconds ....... OK
With what velocity (in m/s) will the arrow hit the moon? ?? (How do i calculate for this?)
1. The speed at the time t is calculated by:

$\displaystyle v(t)=h'(t)=78-1.66t$

Therefore calculate v(94)

2. In this case it is actually not necessary to "calculate" any value because you know that

$\displaystyle v(0) = v(93.9759) = 78\ \frac ms$

4. Thank you vey much. I got it now.