Solve all this calculus question

Here are the solutions to the calculus questions: 1. Evaluate the following limits: b) _x 0 ( 3x)/( 5x) Step 1: Direct substitution yields ((0))/((0)) = (0)/(0), an indeterminate form. Step 2: Apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of 3x is 3 3x. The derivative of 5x is 5 ^2 5x. _x 0 (3 3x)/(5 ^2 5x) Step 3: Substitute x=0 into the new expression. (3 (0))/(5 ^2(0)) = (3 · 1)/(5 · 1^2) = (3)/(5) The limit is (3)/(5). c) _r (3r^3 - 2r)/(r^3 + 5r^2 + 1) Step 1: Divide every term in the numerator and denominator by the highest power of r in the denominator, which is r^3. _r (3r^3)/(r^3) - (2r)/(r^3)(r^3)/(r^3) + (5r^2)/(r^3) + (1)/(r^3) Step 2: Simplify the expression. _r (3 - 2)/(r^2)1 + (5)/(r) + (1)/(r^3) Step 3: Apply the limit. As r , terms of the form (c)/(r^n) approach 0. (3 - 0)/(1 + 0 + 0) = (3)/(1) = 3 The limit is 3. 2. Differentiate from first principles: a) y = x^2 + 3x Step 1: Write the definition of the derivative from first principles. (dy)/(dx) = _h 0 (f(x+h) - f(x))/(h) Step 2: Substitute f(x) = x^2 + 3x and f(x+h) = (x+h)^2 + 3(x+h). f(x+h) - f(x) = [(x+h)^2 + 3(x+h)] - [x^2 + 3x] = [x^2 + 2xh + h^2 + 3x + 3h] - [x^2 + 3x] = 2xh + h^2 + 3h Step 3: Substitute this into the limit and simplify. (dy)/(dx) = _h 0 (2xh + h^2 + 3h)/(h) = _h 0 (h(2x + h + 3))/(h) = _h 0 (2x + h + 3) Step 4: Apply the limit as h 0. (dy)/(dx) = 2x + 0 + 3 = 2x + 3 The derivative is 2x + 3. b) y = x Step 1: Write the definition of the derivative from first principles. (dy)/(dx) = _h 0 (f(x+h) - f(x))/(h) Step 2: Substitute f(x) = x and f(x+h) = (x+h). f(x+h) - f(x) = (x+h) - x Step 3: Use the trigonometric identity A - B = -2 ((A+B)/(2)) ((A-B)/(2)). (x+h) - x = -2 ((2x+h)/(2)) ((h)/(2)) = -2 (x + (h)/(2)) ((h)/(2)) Step 4: Substitute into the limit and rearrange. (dy)/(dx) = _h 0 (-2 (x + h)/(2)) ((h)/(2))h = _h 0 [ -(x + (h)/(2)) · ((h)/(2))(h)/(2) ] Step 5: Apply the limit, using _ 0 ( )/() = 1. (dy)/(dx) = -(x + 0) · 1 = - x The derivative is - x. 3. Find dy/dx: a) y = (x^2 + 1)^5 Step 1: Use the chain rule. Let u = x^2 + 1, so y = u^5. Step 2: Differentiate y with respect to u and u with respect to x. (dy)/(du) = 5u^4 (du)/(dx) = 2x Step 3: Apply the chain rule (dy)/(dx) = (dy)/(du) · (du)/(dx). (dy)/(dx) = 5u^4 · 2x = 5(x^2 + 1)^4 · 2x = 10x(x^2 + 1)^4 The derivative is 10x(x^2 + 1)^4. b) y = x^3 e^-2x Step 1: Use the product rule: (d)/(dx)(uv) = u'v + uv'. Let u = x^3 and v = e^-2x. Step 2: Find the derivatives of u and v. u' = 3x^2 v' = -2e^-2x Step 3: Apply the product rule. (dy)/(dx) = (3x^2)(e^-2x) + (x^3)(-2e^-2x) (dy)/(dx) = 3x^2 e^-2x - 2x^3 e^-2x Step 4: Factor out common terms. (dy)/(dx) = x^2 e^-2x(3 - 2x) The derivative is x^2 e^-2x(3 - 2x). c) y = ( x) Step 1: Use the chain rule. Let u = x, so y = u. Step 2: Differentiate y with respect to u and u with respect to x. (dy)/(du) = (1)/(u) (du)/(dx) = - x Step 3: Apply the chain rule (dy)/(dx) = (dy)/(du) · (du)/(dx). (dy)/(dx) = (1)/( x) · (- x) = -( x)/( x) = - x The derivative is - x. d) y = (2x) Step 1: Use the chain rule. The derivative of u is (1)/(1+u^2) (du)/(dx). Let u = 2x. Step 2: Find the derivative of u with respect to x. (du)/(dx) = 2 Step 3: Apply the chain rule. (dy)/(dx) = (1)/(1+(2x)^2) · 2 = (2)/(1+4x^2) The derivative is (2)/(1+4x^2). e) x^2 y + y^3 = 5 (implicit) Step 1: Differentiate both sides of the equation with respect to x. Remember to use the product rule for x^2 y and the chain rule for y^3. (d)/(dx)(x^2 y) + (d)/(dx)(y^3) = (d)/(dx)(5) Step 2: Apply differentiation rules. (2x)y + x^2 (dy)/(dx) + 3y^2 (dy)/(dx) = 0 Step 3: Group terms containing (dy)/(dx) and solve for (dy)/(dx). (x^2 + 3y^2) (dy)/(dx) = -2xy (dy)/(dx) = (-2xy)/(x^2 + 3y^2) The derivative is (-2xy)/(x^2 + 3y^2). f) x = e^t t, y = e^t t (parametric) Step 1: Use the formula for parametric differentiation: (dy)/(dx) = (dy/dt)/(dx/dt). Step 2: Find (dx)/(dt) using the product rule. (dx)/(dt) = (d)/(dt)(e^t t) = e^t t + e^t (- t) = e^t( t - t) Step 3: Find (dy)/(dt) using the product rule. (dy)/(dt) = (d)/(dt)(e^t t) = e^t t + e^t t = e^t( t + t) Step 4: Substitute into the formula for (dy)/(dx). (dy)/(dx) = (e^t( t + t))/(e^t( t - t)) Step 5: Simplify by canceling e^t. (dy)/(dx) = ( t + t)/( t - t) The derivative is ( t + t)/( t - t). 4. Find and classify all stationary points of y = 2r^3 - 9r^2 + 12r - 3. Step 1: Find the first derivative, (dy)/(dr). (dy)/(dr) = 6r^2 - 18r + 12 Step 2: Set the first derivative to zero to find critical points. 6r^2 - 18r + 12 = 0 Divide by 6: r^2 - 3r + 2 = 0 Factor the quadratic equation: (r-1)(r-2) = 0 The critical points are r=1 and r=2. Step 3: Find the corresponding y values for these critical points. For r=1: y = 2(1)^3 - 9(1)^2 + 12(1) - 3 = 2 - 9 + 12 - 3 = 2. Point: (1, 2). For r=2: y = 2(2)^3 - 9(2)^2 + 12(2) - 3 = 16 - 36 + 24 - 3 = 1. Point: (2, 1). Step 4: Find the second derivative, (d^2y)/(dr^2), to classify the stationary points. (d^2y)/(dr^2) = 12r - 18 Step 5: Evaluate the second derivative at each critical point. For r=1: (d^2y)/(dr^2) = 12(1) - 18 = -6. Since -6 < 0, (1, 2) is a local maximum. For r=2: (d^2y)/(dr^2) = 12(2) - 18 = 6. Since 6 > 0, (2, 1) is a local minimum. The stationary points are: (1, 2), which is a local maximum*. (2, 1), which is a local minimum*. 5. A rectangular box with no lid has a square base of side x and height h. Given that the total surface area is 75 cm^2, find the dimensions for maximum volume. Step 1: Write expressions for the volume (V) and surface area (A). Volume: V = x^2 h Surface Area (base + 4 sides): A = x^2 + 4xh Step 2: Use the given surface area (A=75 cm^2) to express h in terms of x. 75 = x^2 + 4xh 4xh = 75 - x^2 h = (75 - x^2)/(4x) Step 3: Substitute h into the volume equation to get V(x). V(x) = x^2 ((75 - x^2)/(4x)) = (x(75 - x^2))/(4) = (75x - x^3)/(4) = (75)/(4)x - (1)/(4)x^3 Step 4: Find the first derivative of V(x) and set it to zero to find critical points. (dV)/(dx) = (75)/(4) - (3)/(4)x^2 (75)/(4) - (3)/(4)x^2 = 0 75 - 3x^2 = 0 3x^2 = 75 x^2 = 25 Since x must be positive, x = 5 cm. Step 5: Find the corresponding height h. h = (75 - (5)^2)/(4(5)) = (75 - 25)/(20) = (50)/(20) = 2.5 cm Step 6: Use the second derivative test to confirm it's a maximum. (d^2V)/(dx^2) = -(6)/(4)x = -(3)/(2)x At x=5: (d^2V)/(dx^2) = -(3)/(2)(5) = -(15)/(2). Since this is negative, it's a maximum. The dimensions for maximum volume are x = 5 cm and h = 2.5 cm. 6. Water leaks from a conical tank (vertex down) at 3 m^3/min. The tank has radius 2 m and height 5 m. Find the rate at which the water level falls when the depth is 3 m. Step 1: Identify given information and what to find. Given: (dV)/(dt) = -3 m^3/min (negative because water is leaking). Tank dimensions: R = 2 m, H = 5 m. Find: (dh)/(dt) when h = 3 m. Step 2: Write the volume of a cone: V = (1)/(3) r^2 h. Step 3: Use similar triangles to relate r and h for the water in the cone. (r)/(h) = (R)/(H) = (2)/(5) r = (2)/(5)h Step 4: Substitute r into the volume formula to express V in terms of h. V = (1)/(3) ((2)/(5)h)^2 h = (1)/(3) ((4)/(25)h^2) h = (4)/(75)h^3 Step 5: Differentiate V with respect to time t. (dV)/(dt) = (4)/(75) · 3h^2 (dh)/(dt) = (4)/(25)h^2 (dh)/(dt) Step 6: Substitute the known values ((dV)/(dt) = -3, h=3) and solve for (dh)/(dt). -3 = (4)/(25)(3)^2 (dh)/(dt) ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._