ap calculus help. The graph is a parabola

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venialove

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Let be the function given by f(x) = 3 cos x. The graph of f crosses the y-axis at point P and the x-axis at point Q.
Write an equation for the line passing through points P and Q

Point P(0,3)
Point Q(?/2, 0)

This is what I did

y-3=3 cos x(x-0)
y-3= -sin 3x(x^2-0)

This is what I got so far. I plugged in the values then took that derivative of that 3 cos x

Code: 
    |    ,---. 
    |  /       \ y = f(x)
    |/           \
    |P = (0, 3)    \
    |               \
    |                \
    |   Q = (pi/2, 0) \
----+------------------\--
    |                   \
    |
 
I don't understand. What about the 3 cos x. WHere did you the value K from?

Let be the function given by f(x) = 3 cos x. The graph of f crosses the y-axis at point P and the x-axis at point Q.
Write an equation for the line passing through points P (0,3) and Q (?/2, 0).
I thought the values for P and Q had to be set up like this equation y-y1=m(x-x1)
then plug in the value for f(x)= 3 cos x
 
you need to mention that there is a diagram associated with this AP problem ... point Q is the first x-intercept for x > 0.

P(0,3) ... Q(pi/2, 0)

find the slope ... m = (3 - 0)/(0 - pi/2) = -6/pi

use the point-slope form with point P ... y - 3 = (-6/pi)(x - 0)

fyi ... the graph is not a "parabola" , it's a cosine curve.
 
Here's a graph of your 3cos(x) with the tangent line. Graphs always help.

Using y=mx+b and point (0,3), m=-6/Pi

\(\displaystyle 3=\frac{-6}{\pi}(0)+b; \;\ b=3\)

\(\displaystyle y=\frac{-6}{\pi}x+3\)
 

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It says Find the x-coordinate of the point on the graph of f, between points P and Q, at which the line tangent to the graph of f is parallel to the line PQ.
 
What is the slope of the line -- (-6/pi)

What is the slope of the tangent line (3 sin(x))

equate those two and find 'x'.
 
f(x) = 3cos(x), f ' (x) = -3sin(x), y = (-6/Pi)x + 3

-3sin(x) =-6/Pi, x = arcsin(2/Pi) = .690, f(.690) = 2.314, m = -6/Pi = -1.91

y-2.314 = -1.91(x-.690), y = -1.91x+3.6319, the line parallel to the line PQ and tangent to f(x).
 
then it says Let R be the region in the first quadrant bounded by the graph of f and the line segment PQ. Write the integral expression for the volume of the solid generated by revolving the region R about the a-xis. Do not evaluate.

This is what I got
integral -1.91/3cosx
=-sinx 3
f(x)' 1/3 -sin x
 
The line crosses the x-axis at Pi/2. So the limits of integration are 0 to Pi/2.

Using washers:

Then we have \(\displaystyle {\pi}\int_{0}^{\frac{\pi}{2}}\left[(3cos(x))^{2}-(\frac{-6x}{\pi}+3)^{2}\right]dx\)

We can also use 'shells':

\(\displaystyle 2{\pi}\int_{0}^{3}y\left[cos^{-1}(\frac{y}{3})+\frac{\pi(y-3)}{6}\right]dy\)
 
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