Tutor profile: Sara N.
Find the derivative, dy/dx, of the function y = e^x * sin(x).
For this problem we need to take the derivative using the product rule: dy/dx = f'(x) * g(x) + f(x) * g'(x) Let f(x) = e^x and g(x) = sin(x) Then f'(x) = e^x and g'(x) = cos(x) We can then substitute the different pieces into the product rule formula to find the derivative: dy/dx = e^x*sin(x) + e^x*cos(x)
If an object is dropped vertically off of a cliff, and reaches the ground in 10 seconds, how high is the cliff?
For this problem, you'd use one of the kinematic equations for constant acceleration, when acceleration is due to gravity, so you'd use a value of -9.8 m/s^2. The full equation is: d_f = d_i * v_i * t -0.5 * 9.8 *t^2 The initial distance is =0 m, the initial velocity is 0 m/s, and we are told time is = 10 s. We are trying to find the final distance. d_f = 0 + 0*10 - 0.5*9.8*(10)^2 d_f = -0.5* 9.8 * (10)^2 d_f = -4.9 * 100 d_f = -490 m Note that the distance is negative because the object has fallen 490 m below the starting point.
Describe the graph of y = -(x+2)^2 + 5.
This is the vertex form of a parabola, which gives you the vertex and shape/orientation of the graph. The vertex form, for any a, h and k, is y = a(x-h)^2 + k. From this form we know that the vertex is at the coordinate point (h,k). The "a" tells us if the graph is stretched or compressed, as well as if it's open up or down. First, note that the vertex form has a negative sign inside the parentheses. Yet the x-coordinate of the vertex is at the positive value. That means that you always use the opposite value that you see presented. The y - coordinate of the vertex works differently: the base formula has a positive, and the corresponding y-coordinate is also positive, so it keeps the sign that's shown in the equation. The vertex, then, for our problem is (-2, 5), opposite the value we see inside the parentheses for x, and the same value that we see outside the parentheses for y. This means the graph has been shifted 2 units to the left, and five units up. For understanding the effect of the "a" value: * if a > 0 then the graph is compressed toward the y axis, opening upward * if 0 < a < 1 then the graph is stretched away from the y axis, opening upward * if a < 0 then the same rules apply as above, but the graph opens downward Since the a value in our problem is -1, that means that we would only see a reflection of the graph, and the parabola would open downward.
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