A right triangle has base x feet and height h feet, where x is constant and h changes with respect to time t, measured in seconds. The angle θ, measured in radians, is defined by tanθ=hx. Which of the following best describes the relationship between dθdt, the rate of change of θ with respect to time, and dhdt, the rate of change of h with respect to time?
AP CALC AB 4.4 MCQ

A. dθdt=(xx2+h2)dhdt radians per second

B. dθdt=(x2x2+h2)dhdt radians per second

C.dθdt=(1x2+h2√)dhdt radians per second

D. dθdt=tan−1(1xdhdt) radians per second

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s906236 s906236 · Answer 1 · 2022-01-25T19:30:55+01:00

The relationship between dθ/dt, the rate of change of θ regarding time, and dx/dt, the rate of change of x regarding time, is

even the following parameters:

Adjacent side = x

Opposite side = h

The angle of elevation = θ

According to SOH CAH TOA identity;

Substitute the values into the formula

Differentiate regarding x

According to the chain rule,

Hence, the relationship between dθ/dt, the rate of change of θ regarding time, and dx/dt, the rate of change of x regarding time is

msargsya0010 msargsya0010 · Answer 2 · 2022-03-27T18:37:26+02:00

Answer:

x/(x^2 + h^2)

Explanation:

We can set up the equation tan Θ = (h/x)

Since we do not want to deal with a sec^2 Θ (d/dx of tan Θ) its better to move tan to the other side:

tan^-1 Θ * tan Θ = tan^-1(h/x)

Θ=tan^-1(h/x)

Now differentiate:

dΘ/dt = 1/1+(h/x)^2 * d/dx(h/x)

Since x is constant, d/dx = 1/x

We can use a common multiple to change the bottom of the fraction to

dΘ/dx = ( 1 / (x^2+h^2)/x^2 * 1/x )

Rearrange to:

dΘ/dx = x^2/x^2+h^2 * 1/x = x^2 / x(x^2 + h^2)

Now cancel the outside x and the answer is:

dΘ/dx = x/(x^2+h^2)