Answer:
4.6 degrees or zero degree
Step-by-step explanation:
We are given that two curves
[tex]y=2x^2,y=2x^3[/tex]
We have to find the angle between two curves at the point of intersection.
At the point of intersection the value of both curves equal
Therefore, [tex]2x^2=2x^3[/tex]
[tex2x^2-2x^3=0[/tex]
[tex]2x^2(1-x)=0[/tex]
[tex]x=0, x=1[/tex]
Let [tex]g(x)=2x^2[/tex]
[tex]f(x)=2x^3[/tex]
[tex]g'(x)_1=4x[/tex]
[tex]g(1)=m_1=4,g(0)=0[/tex]
[tex]f'(x)=6x^2[/tex]
[tex]f'(1)=m_2=6[/tex]
[tex]f'(0)=0[/tex]
The angle between two curves is given by
[tex]tan\alpha=\mid \frac{m_1-m_2}{1+m_1m_2}\mid [/tex]
Substitute the values then we get
[tex]tan\alpha=\mid \frac{4-6}{1+(4)(6)}\mid [/tex]
[tex]tan\alpha=\frac{2}{25}[/tex]
[tex]\apha=tan^{-1}(\frac{2}{25})=tan^{-1}(0.08)=4.6 degree[/tex]
If we substitute [tex]m_1=m_2=0[/tex]
[tex]tan\alpha=\mid \frac{0-0}{1+0}\mid=0[/tex]
Hence, the acute angles between two given curves=4.6 degrees or 0 degree.
