Question

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

Answer 1

x/(x^2 + h^2)

Step-by-step 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)