Having trouble. Involves rate of change.

**I am not exactly sure how to start. If someone can please help me with one of these word problems I would be extremely grateful.**

An airplane is flying at an altitude of 3 miles at a rate of 400 miles per hour. Find the rate of change of angle a at the moment that the observer is standing at a horizontal distance of 4 miles from the plane,

A ladder 15 ft long is leaning against the wall of a house. The base of the ladder is pulled away from the house at a rate of 2 ft per second. **Consider the triangle created by the ground the wall and the ladder. How fast is the area changing when the ladder’s top is 12 ft up and falling at a rate of 1/2 ft/sec?**

Re: Having trouble. Involves rate of change.

Draw a right angled triangle with height 3, base x and hypotenuse (that is line from observer to plane) making angle A with the base. If the plane is approaching observer dx/dt=-400 We are trying to find dA/dt

Now dA/dt=dx/dt times dA/dx

From the diagram tanA=3/x So x=3cotA dx/dA=-3cosecAcotA So dA/dx=1/(-3cosecAcotA)=-1/3sinAtanA

So dA/dt=-400times -1/3sinAtanA

When x=4 tanA=3/4 and using Pythagoras sinA=3/5 Put in these values to work out dA/dt

If plane is flying away from observer use dx/dt=+400 instead.

Re: Having trouble. Involves rate of change.

Re: Having trouble. Involves rate of change.

Alternative solution for ladder problem (with same notation)

A=1/2xy A^2=1/4x^2y^2

4A^2=x^2y^2

4A^2=(225-y^2)y^2=225y^2-y^4

Differentiate both sides with respect to t

8AdA/dt=450ydy/dt-4y^3dy/dt

When y=12 x=9 A=54 dy/dt=-1/2

Feed in these values and get dA/dt=1.75