Solve speed with the help of f (x) to f ' (x)

[mimi.janson]

I need to solve questions about f (x) to f ' (x)

Homework Statement

hi i have the function f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2

The function explains the realationsship between time and place while a person is running. (can also be called distance and time). Besides i got told that the person runs for three hours

A) First i get asked to write something that shows the speed as function of time.

So i thought of differentiating it, since that will tell me a lot about the slope.

f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2
→
f ' (x)=-32 x^3 + 9,6 x^2 + 0,8 x

B) so my second question was to find out how fast the person runs after 2 hours. Since i know that x is the time and y is the distance i know that i only have to put 2 in the place of x in my function ...i am though not sure if it is in f(x) or f ' (x) but i believe it is in f ' (x) since that was the one telling about the speed.

so i solved it this way

f ' (2)=-32 *2^3 + 9,6 *2^2 + 0,8*2
→
f ' (2)=-256 + 38,4 + 1,6
f '(2) = -216

but there is something wrong here since the speed cannot possibly end up being a negative result ? so i would like someone to show me what i am doing wrong here.


C) last i have a question that asks me to find out where out of the three hours the person runs with the fastest speed ...i don't really understand how i can solve it.

i think i might need to use a second quadratic equation but i don't really understand it so if someone could help me with understanding please

 

[mimi.janson]

Homework Statement

First i would like to say that I am sorry i firstly posted this thread on a wrong place. So i hope i put it right this time

hi i have the function f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2

The function explains the realationsship between time and place while a person is running. (can also be called distance and time). Besides i got told that the person runs for three hours

A) First i get asked to write something that shows the speed as function of time.

So i thought of differentiating it, since that will tell me a lot about the slope.

f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2
→
f ' (x)=-32 x^3 + 9,6 x^2 + 0,8 x

B) so my second question was to find out how fast the person runs after 2 hours. Since i know that x is the time and y is the distance i know that i only have to put 2 in the place of x in my function ...i am though not sure if it is in f(x) or f ' (x) but i believe it is in f ' (x) since that was the one telling about the speed.

so i solved it this way

f ' (2)=-32 *2^3 + 9,6 *2^2 + 0,8*2
→
f ' (2)=-256 + 38,4 + 1,6
f '(2) = -216

but there is something wrong here since the speed cannot possibly end up being a negative result ? so i would like someone to show me what i am doing wrong here.


C) last i have a question that asks me to find out where out of the three hours the person runs with the fastest speed ...i don't really understand how i can solve it.

i think i might need to use a second quadratic equation but i don't really understand it so if someone could help me with understanding please

 

[eumyang]

A) First i get asked to write something that shows the speed as function of time.

So i thought of differentiating it, since that will tell me a lot about the slope.

f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2
→
f ' (x)=-32 x^3 + 9,6 x^2 + 0,8 x

The bolded is not right. It should be -3.2.

but there is something wrong here since the speed cannot possibly end up being a negative result ?

While the answer for f'(2) will be positive, it's possible for the velocity to be negative (speed is the magnitude of velocity). In the case of a person running, he/she would be running backwards.

 

[Ray Vickson]

Homework Statement

hi i have the function f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2

The function explains the realationsship between time and place while a person is running. (can also be called distance and time). Besides i got told that the person runs for three hours

A) First i get asked to write something that shows the speed as function of time.

So i thought of differentiating it, since that will tell me a lot about the slope.

f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2
→
f ' (x)=-32 x^3 + 9,6 x^2 + 0,8 x

B) so my second question was to find out how fast the person runs after 2 hours. Since i know that x is the time and y is the distance i know that i only have to put 2 in the place of x in my function ...i am though not sure if it is in f(x) or f ' (x) but i believe it is in f ' (x) since that was the one telling about the speed.

so i solved it this way

f ' (2)=-32 *2^3 + 9,6 *2^2 + 0,8*2
→
f ' (2)=-256 + 38,4 + 1,6
f '(2) = -216

but there is something wrong here since the speed cannot possibly end up being a negative result ? so i would like someone to show me what i am doing wrong here.


C) last i have a question that asks me to find out where out of the three hours the person runs with the fastest speed ...i don't really understand how i can solve it.

i think i might need to use a second quadratic equation but i don't really understand it so if someone could help me with understanding please

You are confusing *speed* and *velocity*. Speed is the magnitude (absolute value) of velocity, so is always >= 0; however, velocity can be < 0. For example, if I measure positive distances as being to the East of the origin and negative distances to the West, then a positive velocity means traveling towards the East while a negative velocity means traveling towards the West. However, the speedometer in my car will always give a reading >= 0: it measures speed. In your case, if x = time and f(x) = position, then f'(x) = _velocity_, not speed.

You wrote f ' (x)=-32 x^3 + 9,6 x^2 + 0,8 x, which is not correct; it should be f '(x)= -3,2 x^3 + 9,6 x^2 + 0,8 x. Note that f '(2) > 0, but for larger x the velocity would be < 0; for example, look at x = 4.

RGV

 

[xlava]

Don't think of the derivative as "the slope". The derivative of a function f(x) is another function that represents the rate of change of f(x) at any point x. The rate of change is the slope of the tangent line to that graph of f(x) at point x.

Anyway, for part B: Do you know the relationship between the position (x) function and the velocity (v) function? Since your position function seems to be in terms of t, you can find the velocity at time t using the derivative assuming your units check out.

edit: speed cannot be negative, but velocity can, as long as your math is correct, and you represented v(t) as x'(t), you're good. edit 2: and your math is not correct, check that first :)

C: What happens to the position function when its derivative equals 0? What happens to the velocity function when the acceleration function equals 0?

 

[sjb-2812]

f(x) = -0,8 x^4 + 3,2 x^3 + 0,4 x^2
→
f ' (x)=-32 x^3 + 9,6 x^2 + 0,8 x

Right idea, but check this step

 

[sjb-2812]

See also https://www.physicsforums.com/showthread.php?t=575246 (though this is more extensive)

 

[eumyang]

Reported.