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3 years ago

If a recipe uses 5 cups of flour for every 2 cups of sugar, how much sugar is used for every 1 cup of flour?

Mathematics

1 answer:

Aleks04 [339]3 years ago

5 0

2/5 cup of sugar

you divide 5 by 5 to get 1 cup of flour, and you divide 2 by 5 to get 2/5 cup of sugar

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At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

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Step-by-step explanation:

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