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2005 AP CALCULUS AB FREE-RESPONSE QUESTIONS

Consider the differential equation:

Let y = f(x) be the particular solution to this differential equation with the initial condition f(-1) = 2.

(a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes provided in the test booklet.)

(b) Write an equation for the line tangent to the graph of f at x = -1.

(c) Find the solution f(x) to the given differential equation with the initial condition f(-1) = 2.

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04:28

(a) Find the equation of the plane tangent to the graph of $f(x, y)=x^{2} e^{x y}$ at (1,0)(b) Find the linear approximation of $f(x, y)$ for $(x, y)$ near (1,0)(c) Find the differential of $f$ at the point (1,0)

02:03

In Problem:(a) Find the slope of the tangent line to the graph of each function $f$ at the indicated point.(b) Find an equation of the tangent line at the point.(c) Graph $f$ and the tangent line found in $(b)$ on the same set of axes.$$f(x)=4-e^{x} \text { at }(0,3)$$

01:23

In Exercises 15 and 16, (a) sketch an approximate solution of the differential equation satisfying the given initial condition on the slope field, (b) find the particular solution that satisfies the given initial condition, and (c) use a graphing utility to graph the particular solution. Compare the graph with the sketch in part (a). To print an enlarged copy of the graph, go to MathGraphs.com.

$\frac{d y}{d x}=e^{x}-y$$(0,1)$

02:29

In Exercises 15 and 16, (a) sketch an approximate solution of the differential equation satisfying the given initial condition on the slope field, (b) find the particular solution that satisfies the given initial condition, and (c) use a graphing utility to graph the particular solution. Compare the graph with the sketch in part (a). To print an enlarged copy of the graph, go to MathGraphs.com.$$y^{\prime}+\left(\frac{1}{x}\right) y=\sin x^{2}$$$$(\sqrt{\pi}, 0)$$

Transcript

in this problem we have differential equation differential equation which is divide by the X is equals to Vice Square by X -1. Okay, so from the graph a part we have a graph so we have values 21 and to to Okay and so on. So we have to find the values. So first of all, zero square by zero minus one, It’s Square and zero one. So that is equals to zero. Now 1 square by 0 -1 will be minus one. So two squared by 0 -1 is equal to -4. one square is it upon 2 -1 is equal to fun and two square by 2 -1 is equal to four. So we have calculated all the values. Now B is led by Sequels to affects dysfunction And this is an function of two is 3. So divide by the x derivative is equal to Vice Square by X -1. Limit is xy which is 23. So function of 2, 1, is it? So this is equals to nine is 3 square by 2 -1. So this is equals to nine x 1. So divided by D X is equals to nine.…

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