1
$\begingroup$

I am currently doing a little project which involves using a defined function. I set values for two terms, and I took the derivative and squared the two terms. When I attempt to plot it, I do not get anything in the window, nor do I get any errors. Here is what I have for code:

a = ((1/8)*(((0.1*t) - 3)^5)) - 0.5*(((0.1*t) - 3)^3)
b = ((0.1*t) - 3)^2
e = Sqrt[((a')^2) + ((b')^2)]

I receive the following output for the solution of "e"

Sqrt[(Derivative[1][(-0.5 (-3 + 0.1 t)^3 + 1/8 (-3 + 0.1 t)^5)])^2 + (Derivative[1][((-3 + 0.1 t)^2)])^2]

When I go to plot:

Plot[e, {t, 0, 60}, AspectRatio -> 0.5, Frame -> True]

I get a graph with no plot.

Is there something I am doing wrong? Any help would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ Have a look at a specific value, e[1.23] for instance, for a hint. $\endgroup$ – b.gates.you.know.what Apr 19 '18 at 12:18
  • $\begingroup$ When I do that, I just get whatever I got for "e" with a [1.23] stuck at the end of it. $\endgroup$ – Apple Cola Apr 19 '18 at 12:26
  • $\begingroup$ What are the primes (') are suppose to do? You probably want e = Sqrt[(D[a,t]^2) + (D[b,t]^2)]... $\endgroup$ – Henrik Schumacher Apr 19 '18 at 12:45
  • $\begingroup$ Yeah, you were right. I used that prime notation for another project, but I guess it was just a matter of syntax. Thanks! $\endgroup$ – Apple Cola Apr 19 '18 at 13:05
1
$\begingroup$
   a = ((1/8)*(((0.1*t) - 3)^5)) - 0.5*(((0.1*t) - 3)^3)
   b = ((0.1*t) - 3)^2
   e = Sqrt[(D[a, {t, 1}]^2) + (D[b, {t, 1}]^2)]
  Plot[e, {t, 0, 60}, AspectRatio -> 0.5, Frame -> True, PlotStyle ->Blue, FrameStyle -> Directive[Black, Thick]] 

plot

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.