A curve passes through the point (0, 8) and has the property that the slope of the curve at every point P is t?

Question

anonymous · Answer

as the slope is constant, graph is a straight line
y=tx+c
y=8 while x=0
so, 8=c
so, y=tx+8

anonymous · Answer

@antwan , u got it?

kropot72 · Answer

An equation that meets the requirements is
$$y=e ^{x}+7$$
and 
$$t=e ^{x}$$
$$f \prime y=e ^{x}$$

anonymous · Answer

so let that curve be the function y(t). So we also know that
dy/dt=t
dy=tdt
integrating will give us
y+k=t^2/2+c
y=t^2/2+c-k
c-k is just also a constant so we can write that as c' which is a constant
y=t^2/2+c'
now we need to find the actual solution by finding c' by applying the initial condition y(0)=8,
8=(0)^2/2+c'
c'=8
so our curve will be
y=t^2/2+8

anonymous · Answer

its a Differential equation problem

anonymous · Answer

if t is constant, the curve is line.
so: y-0=t(x-8)
is the equation