Question

AP CAL AB HELP!

Answer 1

Answer:

$\frac{dc}{dt}\approx13.8146\text{ km/min}$

Step-by-step explanation:

We know that the plane travels at a constant speed of 14 km/min.

It passes over a radar station at a altitude of 11 km and climbs at an angle of 25°.

We want to find the rate at which the distance from the plane to the radar station is increasing 4 minutes later. In other words, if you will please refer to the figure, we want to find dc/dt.

First, let's find c, the distance. We can use the law of cosines:

$c^2=a^2+b^2-2ab\cos(C)$

We know that the plane travels at a constant rate of 14 km/min. So, after 4 minutes, the plane would've traveled 14(4) or 56 km So, a is 56, b is a constant 11. C is 90+20 or 115°. Substitute:

$c^2=(56)^2+(11)^2-2(56)(11)\cos(115)$

Evaluate:

$c^2=3257-1232\cos(115)$

Take the square root of both sides:

$c=\sqrt{3257-1232\cos(115)}$

Now, let's return to our law of cosines. We have:

$c^2=a^2+b^2-2ab\cos(C)$

We want to find dc/dt. So, let's take the derivative of both sides with respect to t:

$\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2+b^2-2ab\cos(C)]$

Since our b is constant at 11 km, we can substitute this in:

$\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2-(11)^2-2a(11)\cos(C)]$

Evaluate:

$\frac{d}{dt}[c^2]=\frac{d}{dt}[a^2-121-22a\cos(C)]$

Implicitly differentiate:

$2c\frac{dc}{dt}=2a\frac{da}{dt}-22\cos(115)\frac{da}{dt}$

Divide both sides by 2c:

$\frac{dc}{dt}=\frac{2a\frac{da}{dt}-22\cos(115)\frac{da}{dt}}{2c}$

Solve for dc/dt. We already know that da/dt is 14 km/min. a is 56. We also know c. Substitute in these values:

$\frac{dc}{dt}=\frac{2(56)(14)-22\cos(115)(14)}{2\sqrt{3257-1232\cos(115)}}$

Simplify:

$\frac{dc}{dt}=\frac{1568-308\cos(115)}{2\sqrt{3257-1232\cos(115)}}$

Use a calculator. So:

$\frac{dc}{dt}\approx13.8146\text{ km/min}$

And we're done!