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Anni [7]
3 years ago
11

Using calculus, you can find a function’s maximum or minimum by differentiating and setting the result to zero. Do this for equa

tion x=v20gsin2θ0x=v02 sin⁡2θ0, differentiating with respect to θθ, and thus find the maximum range for θθ.
Mathematics
1 answer:
Elenna [48]3 years ago
7 0

Answer:

\dfrac{dx}{d\theta}=\dfrac{2v_0^2\cos2\theta}{g}

x=\dfrac{v_0^2}{g} at \theta=45

Step-by-step explanation:

x=\dfrac{v_0^2\sin2\theta}{g}

Differentiating with respect to \theta means v_0 and g are constants. So we differentiate only the \sin2\theta term.

Assume y=\sin2\theta.

Let u=2\theta.

\dfrac{du}{d\theta}=2

Also, y becomes

y=\sin u

\dfrac{dy}{du}=\cos u

By chain rule,

\dfrac{dy}{d\theta}=\dfrac{dy}{du}\cdot \dfrac{du}{d\theta}

\dfrac{dy}{d\theta}=\cos u\cdot2=2\cos u=2\cos 2\theta

Adding back the constants,

\dfrac{dx}{d\theta} = \dfrac{2v_0^2\cos2\theta}{g}

Equating this to 0,

\dfrac{dx}{d\theta}=\dfrac{2v_0^2\cos2\theta}{g}=0

\cos2\theta=0

2\theta = \cos^{-1}0=90

\theta=45

Using this value of \theta in the expression for x,

x=\dfrac{v_0^2\sin(2\times145)}{g}=\dfrac{v_0^2\sin90}{g}=\dfrac{v_0^2}{g} (since \sin90=1)

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