I am trying to solve a related rates problem, but I am stuck.?

Does dx/dt = 2 or does dy/dt = -2? I'll presume that dx/dt = 2

cos(T) = x / 25

-sin(T) * dT/dt = (1/25) * dx/dt

x = 7

dx/dt = 2

T is acute, so sin(T) = sqrt(1 - cos(T)^2) = sqrt(1 - (x/25)^2) = sqrt(1 - (7/25)^2) = sqrt(576/625) = 24/25

-(24/25) * dT/dt = (1/25) * 2

-24 * dT/dt = 2

dT/dt = -1/12

The angle is decreasing at 1/12 rads per second

Let x be distance of the base of ladder from wall.

The following assumes that dx/dt = 2 ft/s, (i.e. not dy/dt)

The the height of the top of the ladder is (625 – x^2)^(1/2)

You have called the ladder angle at the base a, so,

tan(a) = [(625 – x^2)^(1/2)]/x ………………………(1)

You need da/dt when x = 7 so the related rates are

da/dt = da/dx * dx/dt and that tells us that we need da/dt

a = arctan{[(625 – x^2)^(1/2)]/x} so we need to differentiate that wrt x.

One approach is to make this substitution

Let u = [(625 – x^2)^(1/2)]/x, so that a = arctan(u)

We will use the standard derivative result

da/du = d/du[arctan(u)] = 1/(u^2 + 1) and it is not too hard to see that

u^2 + 1 = (625 – x^2)/x^2 + 1 = 625/x^2, so, inverting that,

da/du = x^2/625 ………………………………………………(2)

But to get da/dx we still need du/dx

du/dx = d/dx[625/x^2 - 1]^(1/2)

du/dx = (1/2)/[625/x^2 - 1]^(1/2) *(-1250/x^3)

du/dx = -625/[x^3 *(625/x^2 – 1)^(1/2)

du/dx = -625/[x^2√(625 – x^2)] …………………….(3)

From (2) and (3) we have da/dx = da/du * du/dx = -1/√(625 – x^2)

The general expression for rate of change of angle a is given by

da/dt = da/dx * dx/dt = -2/√(625 – x^2), and when x = 7 we have

da/dt = -1/12 rad/sec

What is x? Number of seconds, or distance from building?

see https://www.khanacademy.org/math/ap-calculus-bc/de...