The radius of a circle is decreasing at a rate of 4cm/min. How fast are the area and circumference of the circle changing when the radius is 8cm?

When a circular metal plate is heated in an oven, it's radius increases at the rate of 0.02cm/min. At what rate is the plate's area increasing when the radius is 60cm?

A stone is dropped into a pond sending out circular ripples at a rate of 2ft/sec. How fast is the area enclosed by the ripples changing 10 seconds later?

\(80\pi ft^2/sec\)

After dropping a rock in a pond, the ripples form circles where the radius is changing at a rate of 2 meters per second. Calculate the rate of change of the area when the radius is 3 meters.

Two sides of a triangle (not necessarily a right triangle) are 5m and 8m in length and the angle between them is increasing at a rate of 0.06rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is \(\pi/3\) rad.

A 20ft ladder is leaning against a wall. The foot of the ladder is being pulled away from the wall at a rate of 5 ft/sec while the top of the ladder remains on the wall. What is the rate of change in the area beneath the ladder when the foot is 8ft from the wall?

A 13ft ladder is leaning against a house. The ladder slides down the wall at a rate of 3ft/min. How fast is the ladder moving away from the base of the wall when the foot of the ladder is 5ft from the wall? How fast is the area of the triangle changing? How fast is the angle between the ladder and the ground changing?

We have a non-right triangle. Side a, which is not changing, has length 10m and side b, also not changing, has length 15m. However, the angle \(\theta\) between sides a and b is changing at a rate of 0.02rad/sec. How fast is the area of the triangle changing when \(\theta\) is \(\pi/3\)?

\(\text{Area}=(1/2)ab\sin(C)\)

0.7 m²/sec

The side length of a square increases at a rate of 3in/s. How fast are the area and perimeter of the square changing when the side length is 5in?

A screen saver displays the outline of a 3 cm by 2 cm rectangle and then expands the rectangle in such a way that the 2 cm side is expanding at the rate of 5 cm/sec and the proportions of the rectangle never change. How fast is the area of the rectangle increasing when the longer side is 12 cm?

First, we assign variables and get equations for the terms given in the problem statement. It also may help to draw a figure.
Let's label the sides x and y and assign x to the longer side. This means that the side labeled y is expanding at the given rate of \(dy/dt = 5\) cm/sec.

Secondly, we set up the equations based on what they want us to calculate. They ask us for the rate of change of the area, so we need an area equation. Since this is a rectangle, the area is length times width. In terms of our variables, we have \(A=xy\). They tell us to determine \(dA/dt\) when \(x=12\) cm.

Now we can start to solve this equation. Notice that we can't substitute \(x=12\) and \(y=8\) until after we take the derivative since these variables are changing. If we look at the area equation, \(A=xy\), we have two variables. It is certainly possible to take the derivative at this point, we would need to use the product rule and we would end up with quite a complicated expression. It is easier to reduce this equation to one variable before taking the derivative. To do this, we need to know the relationship between x and y. This information was given in the problem statement saying that the rectangle is initially 3 cm by 2 cm. So we can set up a ratio.
\(\displaystyle{ \frac{3}{x} = \frac{2}{y} }\)
Now we need to determine which variable to solve for. Either one will work but, since we need to eventually solve for \(dA/dt\) and we are given \( dy/dt \), it will save time to keep the variable y and get rid of x. So we solve the ratio for x, giving us \(x = 3y/2\). Substituting this into the area equation, we have
\(\displaystyle{ A = xy = \frac{3y}{2}y = \frac{3y^2}{2} }\)

It is time to take the derivative with respect to t (since this is a related rates problem).
\(\displaystyle{ A = \frac{3y^2}{2} ~~~ \to ~~~ \frac{dA}{dt} = {3y}\frac{dy}{dt} }\).

Now we can substitute the given values, \(\displaystyle{ \frac{dy}{dt} = 5 }\). We are given that \(x=12\) but we don't have any x's in the equation. However, we can use the ratio to find the y value, i.e. \(y = 2x/3 = 2(12)/3 = 8\). So now we are ready to calculate the answer.

\(\displaystyle{ \frac{dA}{dt} = {3y}\frac{dy}{dt} = 3(8~cm)(5cm/sec) = 120 }\) cm²/sec

120 cm²/sec