cm 2 /second and the height is 10 cm? Please use

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1 Hillary Lehman Writing Assignment Calculus 151 Summer In calculus, there are many different types of problems that may be difficult for students to comprehend. One type of problem that may be difficult, for any student, is of related rates or rates of change. A reason that students may find related rates to be a hard or difficult section of the curriculum is because students may have trouble conceptualizing what exactly the problem is asking of them. In related rates problems, students have to read carefully, visualize, set up variables, create an equation to relate all variables, differentiate, plug in values and solve. This is a lengthy list for a single problem and is part of the reason why students may feel overwhelmed when it comes to tackling related rates problems. Following the example below, there are five simple steps to help Calculus students solve related rates problems. Suppose that a right circular cylinder has a FIXED radius of r = 5 cm. How fast is the HEIGHT (h) of the cylinder changing at the moment when the SURFACE AREA (A) is increasing at a rate of 90π cm 2 /second and the height is 10 cm? Please use d/dt notation instead of prime notation. [Hint: In the surface area formula for a right circular cylinder, the surface area of the sides of the cylinder is given by the formula 2π rh. How can you account for the area of the top and bottom? (What shape is it?)]

2 1. Read problem carefully and draw a diagram if possible 2. Introduce notation be assigning variables and recording given information Fixed radius: r r = 5 cm Surface Area of a Cylinder (side only): SA SA = 2 π rh SA of a cylinder (top & bottom included): 2 π r 2 + 2π rh Rate of which the SA is increasing: dsa/dt = 90 π cm 2 /second Height: h h = 10 cm Unknown Rate (rate of which the height (h) of the cylinder is changing): dh/dt dh/dt =? Area of a circle: A A = π r 2

3 3. Write an equation that relates the various quantities in the problem SA = 2 π r π rh Surface Area of the cylinder = Area of the top & bottom + Area of side of the cylinder Our blue piece or variable, in the equation, is our Surface Area (SA). We need to know the equation of the SA because the problem wants us to find how fast the height is changing at the moment when our SA is increasing at a rate of 90 π cm 2 /second. We know that when a problem, in Calculus, is wanting us to find the rate of change, of any set of variables, we know that we must find the derivative of the function to help us solve for the rate of change. This is because rate of change is directly related to differentiation because if you have a function, F(x), the rate of change of our function is F (x) or the derivative of the given function. Our red piece or portion of the SA equation is the area (A) of the top and bottom of our cylinder. It is important to remember, when finding the SA of a cylinder, that the top and bottom of the cylinder are not included in the SA formula of the side of a cylinder (2 π rh). Because the top and bottom of the cylinder are not included in the SA formula of the sides, we must separately add on our red piece of the equation to satisfy all portions of our cylinder. What shape is the top and bottom of the cylinder? If you are unsure, look at the picture above and imagine unrolling our cylinder and taking off the top and bottom. The top and bottom will resemble what kind of shape? The top and bottom of our cylinder will look like a circle! This is where we get the red piece of our equation! We know that the area (A) of a circle is: π r 2 but you must include a coefficient of 2, to our red piece or area of a circle, because we have a top and a bottom piece of the cylinder. Our orange piece or portion of our SA equation is the surface area (SA) of the side of the cylinder. It may be difficult to imagine what the diagram of the side of the cylinder would look like because of its 3 dimensional nature. The picture or diagram above, will help you visualize the cylinder rolled out or flattened. By looking at the picture above, we can see that the side of the cylinder rolled out looks like a rectangle! This is helpful because we can now think of the surface area (SA) of the side of the cylinder as a more

4 well known shape: a rectangle! We know the area (A) of a rectangle to be : l x w or the length times the width. The width, in this case, is our height (h) given in the problem: 10 cm. Our length is the distance around a circle (top or bottom of the cylinder) or more formally known as, the circumference. This means that the length of our rolled out rectangle is known as the circumference of a circle: 2 π r. Therefore, the total area of our rectangle will be: 2π rh or our orange piece in the equation above. 4. Use the chain rule, differentiate both sides of the equation and solve for the unknown rate We now have our equation ( SA = 2 πr πrh ) that will provide us with the solution of our unknown rate of dh/dt; or the rate of change, in height, when the SA of the cylinder is increasing at a rate of 90 π cm 3 /second. Our next step in solving this problem is to apply the chain rule to our equation, on both sides. The chain rule, is a formula in Calculus, to compute the derivative that is composed of two or more functions or pieces of an equation. In step 3, you can easily see, by color coding, that we have multiple pieces of our equation; this is why we choose the Chain Rule method to differentiate this equation. Before applying the chain rule method, we can see that there is an r or radius piece in our equation. By using information stated in the problem, we know that our radius r is a fixed or constant variable in our equation. This allows us to plug in our fixed value of r = 5 cm before we differentiate our equation: SA = 2 π (5 2 ) + 2 π (5)h SA = 50 π + 10πh Now that we have our new, simplified equation, we can now differentiate both sides. By applying the Chain Rule, we are left with our new, derived equation: dsa/dt = 10 π dh/dt On the left side of the equation we have a new piece: dsa/dt. This portion of the equation can be read as The rate of change of the SA with respect to time. On the right side of the equation, we are left with the piece 10 π because in the above equation, 50 π is a constant whose derivative is equal to 0 and 10π is allowed to remain in our equation because it was a constant of h. With our 10 π piece, we can see that we have a new variable to our equation : dh/dt. This is the derivative of h and is also our unknown rate or the change in height with respect to time!

5 5. Substitute the given information from the problem into the equation and solve for the unknown rate With our new, derived equation, we have our missing or unknown piece of the equation that we have been hunting for: dh/dt! A student may get confused at this step of the problem because our equation is not in terms of our unknown rate and may forget that we actually do know the value of the other variable, dsa/dt. Our next step should be to plug in all values in which were given to us in the problem. In this case, we need to plug in our value of dsa/dt, or the rate of which our SA of the cylinder is changing with respect to time. dsa/dt = 10 π dh/dt 90 π cm 3 /second = 10 π dh/dt We can now use simple algebra to solve for our unknown rate of dh/dt by dividing 90 π by 10 π. It is crucial to remember your units of measure when solving for the unknown rate. We are then left with our solution for our unknown rate: dh/dt = 9 cm/second; The height of the cylinder is changing at a rate of nine centimeters per second at the moment when the SA of the cylinder is increasing at a rate of 90 π cm 3 /second.