what will be the graph in the equation below? (why is deriv of x=t^2 straight line?)

[Indranil]

As we know If x=t² (distance w.r.t time graph), the slope is increasing or parabola. If we differentiate x with respect to time we get 2t (x= t², x=2t)
Now my question is If, in the distance(x) and time(t) graph, the tangent line is increasing(the first image), then why is 'the straight-line' in the v-t graph(the second image)?

 

[lev888]

Not sure where you see a contradiction.

 

[Indranil]

Not sure where you see a contradiction.

What do you think If x=t², the tangent line is increasing, then what will be the graph if x=2t in the case instant velocity (by differentiating 'x')?

 

[Harry_the_cat]

What do you think If x=t², the tangent line is increasing, then what will be the graph if x=2t in the case instant velocity (by differentiating 'x')?

"What do you think" - that's a bit rude! You haven't really made it clear what you have an issue with.​

The "v" value of the second graph gives the tangent of the first graph for any given t. Since v is increasing (in the second graph), that corresponds with the gradient of the first graph increasing.

 

[HallsofIvy]

Your confusion may come from your poor phrasing- and possibly misunderstanding.

As we know If x=t² (distance w.r.t time graph), the slope is increasing or parabola.

The slope is increasing as t increases and the graph is a parabola. The way your sentence is worded makes it sound as if "the slope is increasing" and "parabola" mean the same thing.

If we differentiate x with respect to time we get 2t (x= t², x=2t)

No, if x= t² then dx/dt= 2t, not x.

Now my question is If, in the distance(x) and time(t) graph, the tangent line is increasing(the first image), then why is 'the straight-line' in the v-t graph(the second image)?

The slope of the tangent line increases, not the line itself. v= dx/dt= 2t is the equation of straight line.

 

[Indranil]

"What do you think" - that's a bit rude! You haven't really made it clear what you have an issue with.​

The "v" value of the second graph gives the tangent of the first graph for any given t. Since v is increasing (in the second graph), that corresponds with the gradient of the first graph increasing.

'"What do you think" - that's a bit rude!' Please don't get me wrong. Here 'what do you think' means just 'opinions'. Please forgive me If I made any mistakes. Here I am learning not only math but also English. Really I appreciate the kind efforts you all are doing for me.

 

[Indranil]

Your confusion may come from your poor phrasing- and possibly misunderstanding.


The slope is increasing as t increases and the graph is a parabola. The way your sentence is worded makes it sound as if "the slope is increasing" and "parabola" mean the same thing.


No, if x= t² then dx/dt= 2t, not x.


The slope of the tangent line increases, not the line itself. v= dx/dt= 2t is the equation of straight line.

Why do 'x= t²' create parabola and why do 'dx/dt= 2t' create straight line? Could you explain, please?

 

[lev888]

Why do 'x= t²' create parabola and why do 'dx/dt= 2t' create straight line? Could you explain, please?

What is the graph of x = t²? What is the graph of v = 2t?

 

[Harry_the_cat]

'"What do you think" - that's a bit rude!' Please don't get me wrong. Here 'what do you think' means just 'opinions'. Please forgive me If I made any mistakes. Here I am learning not only math but also English. Really I appreciate the kind efforts you all are doing for me.

Ok no problem. I read it as "What do you think" with an emphasis on the word think, which has a rather impolite tone to it in the English language. The English language is weird like that. My apologies.

 

[mmm4444bot]

… x= t², x=2t …

… the tangent line is increasing(the first image), then why is 'the straight-line' in the v-t graph(the second image)?

Those two notations are not correct.

2t is the first derivative of t^2, so they are different functions. You're not allowed to name both functions 'x'. You need different names.

x is a function of t

v is a function of t

function v is the first derivative of function x

x(t) = t^2

v(t) = x'(t) = 2t


Next, I'm not sure what you're asking about. Do you have a friend who can translate it for you? Or, maybe translate.google.com will help out.

Are you asking why the graph of 2t is straight?

Are you trying to say you don't understand why lines tangent to x(t) always have slope 2t?

 

[Harry_the_cat]

Why do 'x= t²' create parabola and why do 'dx/dt= 2t' create straight line? Could you explain, please?

Maybe your problem is with the use of the variables.

Do know what the graph of \(\displaystyle y=x^2\) looks like ? Do you know that it is a parabola with its vertex at the origin, if draw on the standard x and y axes?

Do you know that the graph of \(\displaystyle y=2x\) is a straight line through the origin.

The graph of \(\displaystyle x=t^2\) is the same as

\(\displaystyle y=x^2\), except instead of x and y axes, you have t and x axes.

The graph of \(\displaystyle v=2t\) is the same as

\(\displaystyle y=2x\) , except that the axes are t and v.
​

Is that what is troubling you?​