No problem! It's a small but important distinction to make. Glad I could help.

In summary, the height of an object's vertical movement can be described by the equation s = -16t^2 + 96t + 112, where s is measured in feet and t is measured in seconds. The object's velocity can be found by taking the derivative of the equation, which is -32t + 96. When t=0, the object's velocity is 96 m/s. To find the maximum height and the time at which it occurs, set the derivative equal to 0 and solve for t. Plugging this value into the original equation will give the maximum height, which in this case is 256 feet. To find the velocity when the object's height is 0, set the original equation
  • #1
zeldajae
8
0

Homework Statement


The height of an object moving vertically is given by
s = -16t2 + 96t + 112,
when s in ft and t in sec.
a. Find the object's velocity when t=0
b. Find its maximum height and when it occurs.
c. Find its velocity when s=0

The Attempt at a Solution



a.
Now I know to find the objects velocity, take the derivative, which is:

v(t) = s' = -32t + 96.
when t=0, the objects velocity is 96 m/s

b.
This is when I got stuck. When the velocity is 96 m/s, what is its maximum height. I am guessing set the derivative's velocity equal to 0, and solve for t. and plug the value of t in the anti derivative equation, and you would get the maximum height, is the correct? Because whenever the derivative is zero, the anti derivative would either be a max or min, in this case, a max.

0 = -32t + 96
t = 3 sec

s = -16t2 + 96t + 112 plug t= 3 in this equation and you will get,
s = 256 ft?

c.

With the assumptions that my work is so far correct, this is what I would do. set the anti derivative equal to zero and solve for t, using factorization.

0 = -16t2 + 96t + 112
0 = -16(t2 -6t - 7)
0 = -16(t + 1)(t - 7)
t = -1, t = 7, because those are the values when the height is zero.

I would plug those values into the derivative to find the velocity at that point.

v(-1) = 128 m/s
v(7) = -128 m/s

Is all my work correct? Or did I made some mathematical or logical errors?
 
Last edited:
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  • #2
Your answers are correct, except you used the wrong units in a).
Also in b, you say "When the velocity is 96 m/s, what is its maximum height.". That statement is nonsense, and you don't need it. The 96 m/s was about a specific time instance (t = 0), now we're talking about a different time instance (namely, t for which v is maximal) which you find by setting s' = 0.

Finally, I want to make a remark - perhaps a bit picky - about your use of the word "anti derivative". Personally, I would call s "the (distance / given / original) function", and v its derivative. If v was given, I would call that the given/original function and refer to s as the anti-derivative. Now when you say "anti-derivative" it may mean that you are looking at a function f, such that the derivative f' = s (and f'' = s' = v).
 
  • #3
CompuChip said:
Your answers are correct, except you used the wrong units in a).
Also in b, you say "When the velocity is 96 m/s, what is its maximum height.". That statement is nonsense, and you don't need it. The 96 m/s was about a specific time instance (t = 0), now we're talking about a different time instance (namely, t for which v is maximal) which you find by setting s' = 0.

Finally, I want to make a remark - perhaps a bit picky - about your use of the word "anti derivative". Personally, I would call s "the (distance / given / original) function", and v its derivative. If v was given, I would call that the given/original function and refer to s as the anti-derivative. Now when you say "anti-derivative" it may mean that you are looking at a function f, such that the derivative f' = s (and f'' = s' = v).


Thank you so much.

And it was interesting how I never thought about the anti-derivative, like what it may mean when not used appropriately, thank you.
 

What is the definition of "derivative"?

The derivative of a function is the rate of change of that function at a specific point. It represents the slope of the tangent line to the graph of the function at that point.

What are some real-life applications of derivatives?

Derivatives have numerous applications in fields such as physics, economics, engineering, and medicine. Some examples include optimizing production and profit in business, predicting the motion of objects, and finding the maximum and minimum values of a function.

What are the different types of derivatives?

The most common types of derivatives are the first derivative (also known as the derivative or rate of change), second derivative (rate of change of the rate of change), and higher-order derivatives (rate of change of higher order rates). Other types include partial derivatives, implicit derivatives, and logarithmic derivatives.

How do you find the derivative of a function?

The derivative of a function can be found by using differentiation rules, such as the power rule, product rule, quotient rule, and chain rule. These rules provide a systematic way to find the derivative of a function without having to use the limit definition of derivative.

Why are derivatives important in calculus?

Derivatives are essential in calculus because they help us understand the rate of change of a function, which is crucial in many real-life situations. They also allow us to find maximum and minimum values of a function, which is important in optimization problems. Additionally, derivatives are used in many advanced calculus concepts, such as integration and differential equations.

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