let x and y be differentiable functions of t, and let s = \\sqrt{5x^{2}+2y^{2}} be a function of x and y…

let x and y be differentiable functions of t, and let s = \\sqrt{5x^{2}+2y^{2}} be a function of x and y. answer the following questions.\na. how is \\frac{ds}{dt} related to \\frac{dx}{dt} if y is constant?\n\\frac{ds}{dt} = y

let x and y be differentiable functions of t, and let s = \\sqrt{5x^{2}+2y^{2}} be a function of x and y. answer the following questions.\na. how is \\frac{ds}{dt} related to \\frac{dx}{dt} if y is constant?\n\\frac{ds}{dt} = y

Answer

Explanation:

Step1: Differentiate (s) with respect to (t) using chain - rule

Since (s = \sqrt{5x^{2}+2y^{2}}) and (y) is constant, we first rewrite (s=(5x^{2}+2y^{2})^{\frac{1}{2}}). By the chain - rule (\frac{ds}{dt}=\frac{ds}{dx}\cdot\frac{dx}{dt}). Differentiate (s) with respect to (x): (\frac{ds}{dx}=\frac{1}{2}(5x^{2}+2y^{2})^{-\frac{1}{2}}\cdot(10x)).

Step2: Simplify the expression for (\frac{ds}{dx})

(\frac{ds}{dx}=\frac{10x}{2\sqrt{5x^{2}+2y^{2}}}=\frac{5x}{\sqrt{5x^{2}+2y^{2}}}). Then, since (\frac{ds}{dt}=\frac{ds}{dx}\cdot\frac{dx}{dt}), we have (\frac{ds}{dt}=\frac{5x}{\sqrt{5x^{2}+2y^{2}}}\cdot\frac{dx}{dt}).

Answer:

(\frac{5x}{\sqrt{5x^{2}+2y^{2}}}\cdot\frac{dx}{dt})