Answer:
The slope of the curve at (3,1) is 3.
Step-by-step explanation:
The given differential equation is
![y'(t)=t^2-6y^2](https://tex.z-dn.net/?f=y%27%28t%29%3Dt%5E2-6y%5E2)
It is given that the solution curve is passing through the point (3,1).
The slope of a curve y(t) at a point (a,b) is the value of y'(t) at (a,b).
We need to find the slope of the curve at (3,1).
![m=[y'(t)]_{(3,1)}](https://tex.z-dn.net/?f=m%3D%5By%27%28t%29%5D_%7B%283%2C1%29%7D)
![m=[t^2-6y^2]_{(3,1)}](https://tex.z-dn.net/?f=m%3D%5Bt%5E2-6y%5E2%5D_%7B%283%2C1%29%7D)
Substitute t=3 and y=1 in the above equation, to find the slope.
![m=(3)^2-6(1)^2](https://tex.z-dn.net/?f=m%3D%283%29%5E2-6%281%29%5E2)
![m=9-6](https://tex.z-dn.net/?f=m%3D9-6)
![m=3](https://tex.z-dn.net/?f=m%3D3)
Therefore the slope of the curve at (3,1) is 3.