Derivative and Velocity

J

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
POST EDITED

What is the height of a ball (in feet) when the velocity is zero? t is measured in seconds. The ball is being thrown upward (from a cliff).

Velocity is the 1st derivative. Right?

\(\displaystyle y = -16t^{2} + 64t + 200\)

\(\displaystyle y' = -32t + 64\)

The answer is 264. How does it get there?:confused:
 
Last edited:
Jason76 said:
What is the height of a ball (in feet) when the velocity is zero? t is measured in seconds. The ball is being thrown upward.

Velocity is the 1st derivative. Right?

\(\displaystyle y = -16t^{2} + 64t + 200\)

\(\displaystyle y' = -32t + 64\)

The answer is 264. How does it get there?:confused:
Click to expand...

What is the value of y' when the height (y) is maximum?

What is the value of 't', when it (y) reaches maximum?
 
Jason76 said:
The problem says the ball is being thrown upward from a cliff.
Click to expand...

That "really" has nothing to do with the question I asked!

What is the value of the derivative of a function when the function reaches local maximum or minimum?
 
Jason, to find the max or min of a function what is the thing we always do with the first derivative?
 
To find critical numbers (maximum or minimum), you set the 1st derivative to \(\displaystyle 0\) and solve for \(\displaystyle x\). To determine whether or not they are maximum or minimum, then you plug in the critical numbers into the 2nd derivative and note whether they are positive or negative.
 
Jason76 said:
To find critical numbers (maximum or minimum), you set the 1st derivative to \(\displaystyle 0\) and solve for \(\displaystyle x\). To determine whether or not they are maximum or minimum, then you plug in the critical numbers into the 2nd derivative and note whether they are positive or negative.
Click to expand...
Good - Now use that knowledge to solve the problem at hand....
 
So now the main goal is to find the maximum (x coordinate) (which we can locate from the extrema by way of the 2nd derivative test). Next, simply find the y coordinate of the maximum by plugging x into the original equation. That would be the answer.
 
But all of that is more "technical" than you need for this problem. The problem asked "What is the height of a ball (in feet) when the velocity is zero?"

YOU calculated that the velocity was y'= -32t+ 64. When will that be 0? What is the height for that value of t?
 
HallsofIvy said:
But all of that is more "technical" than you need for this problem. The problem asked "What is the height of a ball (in feet) when the velocity is zero?"

YOU calculated that the velocity was y'= -32t+ 64. When will that be 0? What is the height for that value of t?
Click to expand...

If you plugin \(\displaystyle 0\) into the derivative, then you get \(\displaystyle 64\) as the value of \(\displaystyle t\) at \(\displaystyle 0\), so the maximum \(\displaystyle y\) coordinate value minus \(\displaystyle 64\) must be the final answer (I think).

But this could be wrong. Perhaps they mean set the derivative to \(\displaystyle 0\) and solve for \(\displaystyle t\). In that case, \(\displaystyle t = 0.5\)
 
Last edited:
Jason76 said:
POST EDITED

What is the height of a ball (in feet) when the velocity is zero? t is measured in seconds. The ball is being thrown upward (from a cliff).

Velocity is the 1st derivative. Right?

\(\displaystyle y(t) = -(16\ ft/s^2)\ t^{2} + (64\ ft/s)\ t + (200\ ft)\)

\(\displaystyle y\prime (t) = -(32\ ft/s^2)\ t + (64\ ft/s)\)

The answer is 264 ft. How does it get there?:confused:
Click to expand...
Jason76 said:
If you plugin [FONT=MathJax_Main]0[/FONT] into the derivative, then you get [FONT=MathJax_Main]64[/FONT] as the value of [FONT=MathJax_Math]t[/FONT] at [FONT=MathJax_Main]0[/FONT],
NO - plug in \(\displaystyle y\prime = 0\) , NOT \(\displaystyle t = 0\)!!!
so the maximum [FONT=MathJax_Math]y[/FONT] coordinate value minus [FONT=MathJax_Main]64[/FONT] must be the final answer (I think).

But this could be wrong. Perhaps they mean set the derivative to [FONT=MathJax_Main]0[/FONT] and solve for [FONT=MathJax_Math]t[/FONT]. In that case, [FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0.5[/FONT]
Click to expand...
STARTING OVER FROM SCRATCH

\(\displaystyle y\) has dimensions of "ft," so every term in any equation for \(\displaystyle y\) MUST have dimensions of "ft."

\(\displaystyle y\prime\) has dimensions of "ft/s," so every term in any equation for \(\displaystyle y\prime\) MUST have dimensions of "ft/s."

Let \(\displaystyle t_1\) be the time at which the velocity has dropped to zero. That will be the peak of the trajectory, since the ball is moving neither up nor down when the velocity is 0.

\(\displaystyle y\prime = -(32\ ft/s^2)\ t_1 + (64\ ft/s) = 0\)

\(\displaystyle \implies t_1 = (64 ft/s)/(32 ft/s^2) = 2\ s\)

What is the value of \(\displaystyle y(t)\) when \(\displaystyle t=t_1\)?

\(\displaystyle y(t_1) = -(16\ ft/s^2)\ (2\ s)^{2} + (64\ ft/s)\ (2\ s) + (200\ ft)\)

.........\(\displaystyle = (-64\ ft) + (128\ ft) + (200\ ft) = 264\ ft\)
 
DrPhil said:
STARTING OVER FROM SCRATCH

\(\displaystyle y\) has dimensions of "ft," so every term in any equation for \(\displaystyle y\) MUST have dimensions of "ft."

\(\displaystyle y\prime\) has dimensions of "ft/s," so every term in any equation for \(\displaystyle y\prime\) MUST have dimensions of "ft/s."

Let \(\displaystyle t_1\) be the time at which the velocity has dropped to zero. That will be the peak of the trajectory, since the ball is moving neither up nor down when the velocity is 0.

\(\displaystyle y\prime = -(32\ ft/s^2)\ t_1 + (64\ ft/s) = 0\)

\(\displaystyle \implies t_1 = (64 ft/s)/(32 ft/s^2) = 2\ s\)

What is the value of \(\displaystyle y(t)\) when \(\displaystyle t=t_1\)?

\(\displaystyle y(t_1) = -(16\ ft/s^2)\ (2\ s)^{2} + (64\ ft/s)\ (2\ s) + (200\ ft)\)

.........\(\displaystyle = (-64\ ft) + (128\ ft) + (200\ ft) = 264\ ft\)
Click to expand...

Looks good to me

What is the height of a ball (in feet) when the velocity is zero? t is measured in seconds. The ball is being thrown upward (from a cliff).

Velocity is the 1st derivative. Right?

\(\displaystyle y = -16t^{2} + 64t + 200\)

\(\displaystyle y' = -32t + 64\)
Click to expand...
 
Last edited:
Jason76 said:
If you plugin \(\displaystyle 0\) into the derivative, then you get \(\displaystyle 64\) as the value of \(\displaystyle t\) at \(\displaystyle 0\), so the maximum \(\displaystyle y\) coordinate value minus \(\displaystyle 64\) must be the final answer (I think).

But this could be wrong. Perhaps they mean set the derivative to \(\displaystyle 0\) and solve for \(\displaystyle t\). In that case, \(\displaystyle t = 0.5\)
Click to expand...
If you really believe this then you have serious algebra and arithmetic problems. I am hoping you just did that too quickly and with a little thought will realize that the solution to -32t+ 64= 0 is NOT t= 1/2.

What is really worrisome is that you are repeatedly posting problem where you are, basically, substituting numbers into formulas without any regard to what they mean. What does "t= 0" mean in this problem?
 
You must log in or register to reply here.
Top