Advanced Placement Calculus AB Practice

Section II Part Answers

A graphing calculator is not allowed for the following problems.

1.Write an equation for the line tangent to the curve x² + 4x + y³ = 4 at the point (1, -1).

Solution: Treating y as an implicitly defined function of x and differentiating both sides of the equation with respect to x, we obtain:
2x + 4 + 3y²y' = 0
y' = -(2x + 4)/(3y²)
At point (1,-1), y'(1) = -2
y = -1 - 2·(x-1)

2. dy/dx = (x + 1)²y
Sketch a slope field for the given differential equation at the points indicated.

Solution:

3. dy/dx = (x + 1)²y
Let y = f(x) be the solution of this equation with the initial condition f(1) = 1. What is the equation of the tangent to the graph of f at x = 1?

Solution: f '(1) = (1 + 1)²1 = 4
The tangent equation is y = 1 + 4(x - 1), or
y = 4x - 3.

4. dy/dx = (x + 1)²y
Find the solution y = f(x) to the given differential equation with the initial condition f(1) = 1.

Solution: dy/dx = (x + 1)²y
(dy/dx)/y = (x + 1)²

ln(y(x)) - ln(y(1)) = x³/3 + x² + x - (1/3 + 1 + 1).
ln(y) = x³/3 + x² + x - 7/3.
y = e^{(x³/3 + x² + x - 7/3)}

5. Consider the closed curve in the xy plane given by 2x² + 5x + y³ + 4y² + y = 8.
Show that dy/dx = -(4x + 5)/(3y² + 8y + 1).

Solution: Treating y as an implicitly defined function of x and differentiating both sides of the equation with respect to x, we obtain:
4x + 5 + 3y²y' + 8yy' + y' = 0
dy/dx = -(4x + 5)/(3y² + 8y + 1)

6.
ap, calculus, graph, function
The graph of function f is shown above. Let:

Calculate g(3). Does g have a relative minimum or a relative maximum at x = -1? Calculate g"(0).

Solution:

g(3) is the area of the pentagon in the first quadrant bounded by the x axis, the y axis and function f.
g(3) = 3/2 + 2 + 1 = 4.5

g'(x) = f(x)
g'(-1) = 0
g'(x) = f(x) < 0 for x < -1 and g'(x) = f(x) > 0 for x > -1 so g has a relative minimum at x = -1.

g"(0) = f '(0) = (2 - 1)/(1 - 0) = 1
g"(0) = 1

7. Let f be the function given by f(x) = (2x² + 5x -1)⁷.
Write the equation for the line tangent to f(x) in x = 0.

Solution: Let u(x) = (2x² + 5x -1)
Then, f '(x) = (du⁷/du)d(2x² + 5x -1)/dx = 7u⁶(4x + 5) = 7(4x + 5)(2x² + 5x -1)⁶
f '(0) = 35 and f(0) = -1.
The line of slope 35 that passes through (0,-1) is y = 35x - 1

8. Find the coordinates of the two points on the x² - 2x + 4y² + 16y + 1 = 0 closed curve where the line tangent to the curve is vertical.

Solution: Treating y as an implicitly defined function of x and differentiating both sides of the equation with respect to y, we obtain:

2xx' - 2x' + 8y + 16 = 0
x'(2x - 2) = -8(y + 2)
x' = -4(y + 2)/(x - 1) and x' should be 0 if the line tangent is vertical.
y = -2
At the points with vertical tangents, x² - 2x + 16 - 32 + 1 = 0
x² - 2x - 15 = 0
The solutions are x₁ = 5 and x₂ = -3.
The coordinates of the two points are (5, -2) and (-3, -2).

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