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Question 2
Figure 3 shows a sketch of part of the curve C with equation y = 3^x The point P lies on C and has coordinates (2, 9). The line l is a tangent to C at P. The line... show full transcript
Step 1
Answer
To find the x-coordinate of point Q, we begin with the equation for the curve:
[ y = 3^x ]
At point P, with coordinates (2, 9), we can confirm that:
[ y = 3^2 = 9 ]
Next, we find the slope of the tangent line at P by differentiating the equation of the curve:
[ \frac{dy}{dx} = 3^x \ln(3)]
Substituting ( x = 2 ), we find the slope (m) at P:
[ m = 3^2 \ln(3) = 9 \ln(3) ]
The equation of the tangent line l at point P can be expressed in point-slope form:
[ y - 9 = 9 \ln(3)(x - 2) ]
To find the x-coordinate of point Q, we set y = 0 (the x-axis):
[ 0 - 9 = 9 \ln(3)(x - 2) ]
Solving for x:
[ -9 = 9 \ln(3)(x - 2) ] [ x - 2 = -\frac{1}{\ln(3)} ] [ x = 2 - \frac{1}{\ln(3)} ]
Thus, the x-coordinate of Q is:
[ x_Q = 2 - \frac{1}{\ln(3)} ]
Step 2
Answer
To find the volume generated by rotating region R about the x-axis, we will use the formula:
[ V = \pi \int_{a}^{b} (f(x))^2 , dx ]
Here, the function describing the curve is ( f(x) = 3^x ). We first determine the limits of integration, which are from ( x = 0 ) to ( x = 2 - \frac{1}{\ln(3)} ) (the x-coordinate found earlier).
Thus, the volume becomes:
[ V = \pi \int_{0}^{2 - \frac{1}{\ln(3)}} (3^x)^2 , dx ] [ = \pi \int_{0}^{2 - \frac{1}{\ln(3)}} 9^{x} , dx ]
Calculating the integral:
[ V = \pi \left[ \frac{9^x}{\ln(9)} \right]_{0}^{2 - \frac{1}{\ln(3)}} ] [ = \pi \left[ \frac{9^{2 - \frac{1}{\ln(3)}} - 9^{0}}{\ln(9)} \right] ] [ = \pi \left[ \frac{9^2 \cdot 9^{-\frac{1}{\ln(3)}} - 1}{\ln(9)} \right] ] [ = \pi \left[ \frac{81 \cdot \frac{1}{3} - 1}{\ln(9)} \right] ] [ = \frac{80 \pi}{3 \ln(9)} ]
Thus, expressing the answer in the suitable fraction form:
[ V = \frac{80\pi}{3\ln(9)} ]
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