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These are questions asked by Tima222

Graph the real roots of the equation x3 + (a − 3)3x2 − a2x + a3=0 for a ∈ [0, 1].



for i from 1 to 3 do


print(plot(R||i(a),a=0..1,numpoints=500)); end od:


Is there a simplier way to solve this. If not why did they choose this path?



what does this mean?

How do I find where 𝑓(π‘₯) = (2*π‘₯ ^2) tan(2*π‘₯), is discontinuous in the interval [0,2πœ‹], and find the discontinuities. I know you need ot use the commands: discount(f(x),x), iscont(f(x),x=a..b,’open’), iscont(f(x),x=a..b,’close’) and fdiscont(f(x),x=a..b, resolution) which will help me find the list of ranges, each of width resolution, in which there appears to be a discontinuity in the function or its first derivative.However, what is next ?

Solve the following ODE for given initial conditions both analytically and numerically. Use the odeadvisor in DEtools package to classify the equations. Plot the explicit and numerical solutions in the range [0,20] using plot and odeplot commands. 𝑦 ′′(𝑑) + 16𝑦(𝑑) = 3 sin(πœ” 𝑑), for πœ” ∈ {1,4,5,8}, 𝑦(0) = 𝑦 ′ (0) = 0


I started by:


ode := diff(y(t), t, t) + 16*y(t) = 3*sin(w*t);
                  /  2      \                       
                  | d       |                       
           ode := |---- y(t)| + 16 y(t) = 3 sin(w t)
                  |   2     |                       
                  \ dt      /                       
odeadvisor(ode, y(t));
            [[_2nd_order, _linear, _nonhomogeneous]]
I undertsnad how to plot 

but how do I odeplot and how do I contribute πœ” ∈ {1,4,5,8

. Let 𝑔(π‘₯) = cos2 (π‘₯ 2+1) (sinπ‘₯+1) 2 and 𝑓(π‘₯) = (2π‘₯ 2 − 1) 3√π‘₯ + 2.

Write a procedure to return the tangent line of the function 𝑓(π‘₯) at the given point π‘₯ ∈ 𝐷𝑓.


I started by writing a regular Proc template 

such as 

tangent line:=proc(y);

however i am unsure if I shpuld go forwrad with using an if loop or not?



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