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Find the domain and range of the function that is the solution to the differential equation $$\frac{dy}{dx} = e^{x+y}$$ and whose graph passes through the origin - HSC - SSCE Mathematics Extension 1 - Question 14 - 2024 - Paper 1

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Find-the-domain-and-range-of-the-function-that-is-the-solution-to-the-differential-equation-$$\frac{dy}{dx}-=-e^{x+y}$$-and-whose-graph-passes-through-the-origin-HSC-SSCE Mathematics Extension 1-Question 14-2024-Paper 1.png

Find the domain and range of the function that is the solution to the differential equation $$\frac{dy}{dx} = e^{x+y}$$ and whose graph passes through the origin. (... show full transcript

Worked Solution & Example Answer:Find the domain and range of the function that is the solution to the differential equation $$\frac{dy}{dx} = e^{x+y}$$ and whose graph passes through the origin - HSC - SSCE Mathematics Extension 1 - Question 14 - 2024 - Paper 1

Step 1

Show that for \theta < \sin^{-1}(\frac{8}{9}), the distance D(t) is increasing for all t > 0.

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Answer

The distance from the origin is given by: D(t)=r(t)=(Vcosθgt)2+(Vsinθggt22)2.D(t) = |\mathbf{r}(t)| = \sqrt{\left(\frac{V \cos \theta}{g} t\right)^2 + \left(\frac{V \sin \theta}{g} - \frac{gt^2}{2}\right)^2}.

To show that D(t) is increasing, we compute the derivative: D(t)=ddt(x2+y2)=xdx/dt+ydy/dtx2+y2.D'(t) = \frac{d}{dt}(\sqrt{x^2 + y^2}) = \frac{x \cdot dx/dt + y \cdot dy/dt}{\sqrt{x^2 + y^2}}.

Consequently, we will show that (D'(t) > 0) for all ( t > 0) when (\theta < \sin^{-1}(\frac{8}{9})) holds true, leading to increasing distance.

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