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Numerade Educator



Problem 11 Easy Difficulty

Explain why the functions with the given graphs can't be solutions of the differential equations

$ \frac {dy}{dt} = e^t (y - 1)^2 $


(a) This function is increasing and also decreasing. But $d y / d t=e^{t}(y-1)^{2} \geq 0$ for all $t$, implying that the graph of the solution of the differential equation cannot be decreasing on any interval.
(b) When $y=1, d y / d t=0,$ but the graph does not have a horizontal tangent line.


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Video Transcript

Hi, I'm gonna solve Question 11 from section 9.1 off. Str calculus This question. As for us to show that these two graphs cannot be a solution, I've given differential equation for solving these questions. We don't need to solve the differential equation, and we just can use the properties off the first derivative. If you have a graph of a function, what does it mean? The first derivative at each point to that growth? If you consider the point off the graph, for example, Mexico's Toe one the derivative of the function at Mexico's toe one is the slope off a tangent line. At that point of the graph, the Tangent line is a straight line, which touches the graph. That's point here. The slope of a tangent line is positive. It means that the derivative of this function and x a close toe one is positive. Now let's return to the differential equation. We know that it's the poverty is always a positive function and also by minus one Squared is always positive. It means that the derivative off the function at each point is always a positive value, So the tangent line at each point off the solution off. This differential equation has a positive the slope, but in grab it is not the case. If we look at, for example, this point, the tangent line is decreasing means that the slope off a tangent to learn is negative. In other wars, the derivative off dysfunction is negative at this point, so graph a cannot be a solution off these differential equations. What about graph B? Again, we return to the differential equation, considering we're equals toe one at this point de vie to DT or that their innovative off the function equal zero. It means that if you have a graph off the function, the tangent line at why equals one at some point. T should be a horizontal line. But if you consider the graph B at what goes toe one in some tea year, the Tangent line has a positive years ago, and it is not horizontal. It means that the derivative of this function is not zero at she not and one so graph. We cannot be a solution off this different John equation