Consider the differential equation \(\displaystyle{\frac{dy}{dt} = t-2}\text{.}\)
Suppose that \(y(t)\) is a solution to this differential equation corresponding to the initial value \(y(0) = 1\text{.}\) On the graph of \(y\) versus \(t\text{,}\) what is the slope of the tangent line to the curve at the point \((0,1)\text{?}\)
Slope =
Suppose (instead) that \(y(t)\) is a solution to this differential equation corresponding to the initial value \(y(0) = 5\text{.}\) On the graph of \(y\) versus \(t\text{,}\) what is the slope of the tangent line to the curve at the point \((0,5)\text{?}\)
Slope =
Suppose that \(y(t)\) is a solution to this differential equation that passes through the point \((248, 142)\text{.}\) On the graph of \(y\) versus \(t\text{,}\) what is the slope of the tangent line to the curve at the point \((248,142)\text{?}\)
Slope =
If we only know the differential equation, (but not any points on the solution curve), which of the following could be a solution to the equation? Select all that apply.
\(\displaystyle y = t^2/2 - t\)
\(\displaystyle y = t^2/2 - 2t + 1\)
\(\displaystyle y = \frac{1}{2}(t-2)^2\)
\(\displaystyle y = t^2/2 - 2t + 142\)
None of the above
Given this differential equation (but not an initial value or the solution) what information is sufficient to determine the slope of the tangent line to the solution curve at a point on the curve?
The slope of the solution curve at \(t=0\text{.}\)
The \(y\)-coordinate of the point.
The \(t\)-coordinate of the point.
None of the above