Calculus Answer Note #1

Question 1 Given the function f(x) = 2x^3 -1 Find the average rate of change of f from x = -4 to x = 2 Give your answer as an integer or reduced fraction. Answer To find the average rate of change of a function, we use the formula (f(b) - f(a)) / (b - a), where a and b are the two given points on the x-axis. In this case, a = -4 and b = 2. Therefore, the average rate of change of f(x) from x = -4 to x = 2 can be calculated as: [f(2) - f(-4)] / (2 - (-4)) = [2(2)^3 - 1 - 2(-4)^3 - 1] / (2 - (-4)) (using the given function f(x) = 2x^3 -1) = [16 - 1 - (-128) - 1] / (2 + 4) = (16 - 129) / 6 = -113/6 Hence, the average rate of change of f(x) from x = -4 to x = 2 is -113/6 or -18.83 units per unit of x. Question 2 Given the function f(x) = -3x^3 +2 Find the average rate of change of f from x=-3 to x=2 Answer Question: Given the function f(x) = -3x^3 +2, find the average rate of change of f from x=-3 to x=2. Response: The average rate of change of a function is defined as the change in output over the change in input. Mathematically, it can be represented as (f(x2) - f(x1))/(x2 - x1), where x1 and x2 are the initial and final values of the input respectively. In this case, x1 = -3 and x2 = 2. We can substitute these values in the formula to get (f(2) - f(-3))/(2 - (-3)) = ((-3(2)^3 + 2) - (-3(-3)^3 + 2))/(5) = (-24 + 2 - 81 + 2)/5 = -101/5 = 20.2 Therefore, the average rate of change of f from x=-3 to x=2 is -20.2. Question 3 Show that f(x) = x2/5 is differentiable at x = 0, then find f’(0). Answer According to Calculus, a function is said to be differentiable at a particular point if its derivative exists at that point. In simpler terms, a function is differentiable at a point if its graph has a tangent line at that point. In order to show that f(x) = x^2/5 is differentiable at x = 0, we have to prove that the derivative of the function exists at x = 0. To find the derivative of f(x), we can use the power rule of differentiation, which states that for any real number n and any differentiable function g(x), d/dx (g(x)^n) = n*g(x)^(n-1)*g'(x) Using this rule, we can find the derivative of f(x) as: f'(x) = (2/5)*(x^(2/5 - 1)*(x)') = (2/5)*(x^(-3/5)) = 2x^(-3/5)/5 Now, to find the derivative at x = 0, we substitute the value of x in the above equation as: f'(0) = 2*0^(-3/5)/5 = 0/5 = 0 Hence, we can see that the derivative of f(x) exists at x = 0, and therefore, we can conclude that f(x) = x^2/5 is differentiable at x = 0. Question 4 Find the volume of the solid E whose boundaries are given in rectangular coordinates. (x, y, z) | sqrt( x2 + y2) ≤ z ≤ sqrt(64 − x2 − y2) , x ≥ 0, y ≥ 0 Answer To find the volume of this solid E, we will use the concept of double integration in Calculus. The region enclosed by the boundaries is a cylindrical shape with a coneshaped cavity at the top. We will use this information to set up our limits and integrals for finding the volume. Firstly, let's visualize the boundaries given to us in rectangular coordinates. From the equations, we can see that the solid E is bounded by a circular base with a radius of 8 units, centered at the origin. The height of the solid varies from the base, starting at the cone-shaped cavity at the top (z = sqrt( x^2 + y^2)), and extending down to the circular base (z = sqrt(64 − x^2 − y^2)). To approach this problem, we will use the cylindrical coordinates and rewrite the equations in terms of r (radius), θ (angle), and z (height). This will help us to easily set up the limits for the double integration. Converting the given equations in cylindrical coordinates, we get: z = r (height varies from 0 to r) z = √(64 - r^2) (height varies from √(x^2 + y^2) to √(64 - x^2 - y^2)) r ranges from 0 to 8 units (radius of the circular base) θ ranges from 0 to 2π radians (we need to integrate the entire circular base) Now that we have set up our limits, we can proceed with the double integration. The volume (V) of the solid is given by the double integral of the function 1 with respect to z, r, and θ. V = ∫∫∫1 dzdrdθ Where the outermost integral is with respect to θ, the middle integral is with respect to r, and the innermost integral is with respect to z. Substituting the limits, we get: V = ∫∫∫1 dzdrdθ = ∫_0^2π∫_0^8∫_0^(√(64 - r^2)) 1 dzrdrdθ = 2π∫_0^8∫_0^(√(64 - r^2)) r dzdr = 2π∫_0^8 r(√(64 - r^2)) dr = 2π (∫_0^8 64^1/2 dr - ∫_0^8 r^3/2dr) = 2π ( 64^3/2/3 - 8^5/5) = 2π (2048/3 - 32768/3) = 2π (-30720/3) = -20480π Hence, the volume of the solid E is -20480π cubic units. Question 5 Evaluate the following integral using a change of variables: RR e^xydA, where D is the region in the first D quadrant bounded by the hyperbolas xy = 1 and xy = 4, and the lines y/x = 1 and y/x = 3. Sketch both the original and new region of integration, D and D∗ Answer To evaluate this integral using a change of variables, we will use the transformation u = xy, v = y/x. This will help us simplify the integral and make it more manageable to solve. Let's break down the process step-by-step. Step 1: Understanding the region of integration First, let's sketch the given region D in the xy-plane. We can see that it is a rectangular region bounded by the hyperbola xy = 1 and xy = 4, and the lines y/x = 1 and y/x = 3. This can be seen in the diagram below. Now, let's also sketch the new region D∗ in the uv-plane. We know that the transformation u = xy, v = y/x will map our original region D to the new region D∗ . The transformation can be visualized as a stretching in both the x and y directions. The diagram below shows the new region of integration D∗ . Step 2: Setting up the integral Now, let's use the transformation to express dA in terms of du and dv. We know that dA = dx dy and since u = xy, we can write dx dy = d(u/v)dv. Substituting this into our integral, we get I = RR e^xydx dy = RR(e^u/v)(1/v)dudv. Step 3: Determining the limits of integration To determine the limits of integration for the new integrals, we need to express the original ones in terms of u and v. Let's start with the equations to define the lines y/x = 1 and y/x = 3. y/x = 1 --> xy = 1 --> u = 1 y/x = 3 --> xy = 3 --> u = 3 Similarly, for the hyperbolas xy = 1 and xy = 4, we get xy = 1 --> u = 1 xy = 4 --> u = 4. From these, we can see that the limits for u are from 1 to 4, and for v, they are from 1 to 3. Step 4: Evaluating the integral Now, let's plug in the limits of integration and the expression for dA in terms of du and dv into our integral. This results in I = RR(e^u/v)(1/v)dudv = RR (e^u/v)/v du dv = R3⌠ 1 4 (e^u/v)/v du dv = R3⌠ 1 4 e^u/v du ⌠ 3 1 1/v dv = 1/v^2⌠ 1 4 e^u/v du = v⌠ 1 4 e^u/v du = v(e^u/v)⌠ 1 4 du = vu(e^u/v)|^4_1 = v(e^4/v - e^1/v). Step 5: Putting it all together Finally, we substitute the value back into our original integral to get the final answer I = RR e^xydA = RR (e^u/v)(1/v)dudv = v(e^4/v - e^1/v) = 3(e^3 - 1) Hence, the value of the given indefinite integral is 3(e^3 - 1).