Point P in the curve y=x^3 has coordinates (3,27) and PQ is the tangent to the curve at P.Point Q touches the x-axis.Find the area of the region enclosed between the curve, PQ and the x-axis.
My answer:
I used differenttiation to find the gradient of the tangent: dy/dx=3x^2.
when x=3, the gradient is 3(3)^2=27
Then I used y=mx+c to find the equation of PQ,so 27=27(3)+c and got that the equation is y=27x-54. Finally, I put y=0 in the equation to find x at point Q and got x=2. I intergrated y=x^3 using the limits 2 and 0 and got 4 units^2 for the area. But the answer should be 27/4???? Could you please help me? Thank you
You cannot just integrate on [0,2] to get the area. That ignores the part between the curve and the line on the interval [2,3]. That is
∫[2,3] x^3-(27x-54) dx = 11/4
Add that to your area of 4, and you get 27/4
The area could also be considered using horizontal strips. In that case, each strip lies between the line and the curve, with no break in the boundary. Then the area is
∫[0,27] (y/27 + 2)-∛y dy = 27/4
To find the area enclosed between the curve, PQ, and the x-axis, you need to integrate the absolute value of the function y = x^3 between the x-values where the curve intersects the x-axis.
Let's go through the steps to find the correct area calculation:
1. First, you correctly found the equation of the tangent line PQ as y = 27x - 54.
2. Next, you need to find the x-coordinate of the point where PQ touches the x-axis. You set y = 0 in the equation of PQ:
0 = 27x - 54
27x = 54
x = 2
So the x-coordinate of the point Q where PQ touches the x-axis is indeed x = 2.
3. Now, you want to find the x-values where the curve y = x^3 intersects the x-axis. To do this, you set y = 0 in the equation of the curve:
0 = x^3
x = 0
Therefore, the point P on the curve is (0, 0).
4. Finally, to find the area between the curve, PQ, and the x-axis, you need to integrate the absolute value of y = x^3 between the x-values 0 and 2:
Area = ∫[0 to 2] |x^3| dx
Since the function x^3 is non-negative on [0, 2], the absolute value is unnecessary. You can simply integrate x^3 from 0 to 2:
Area = ∫[0 to 2] x^3 dx
Evaluating this integral:
Area = [x^4/4] [0 to 2]
Area = (2^4/4) - (0^4/4)
Area = 16/4
Area = 4 units^2
Therefore, the correct area enclosed between the curve, PQ, and the x-axis is indeed 4 units^2, as you initially calculated, not 27/4.
To find the correct area, we need to approach the problem differently.
Let's start by finding the equation of the tangent line PQ.
1. We know that the slope of the tangent line at point P is equal to the derivative of the curve at P.
So, the slope of the tangent line is: m = dy/dx = 3(3)^2 = 27.
2. Now, we have the slope and a point on the line (P with coordinates (3,27)).
We can use the point-slope form of a line to find the equation of PQ.
The point-slope form is: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values, we get: y - 27 = 27(x - 3).
Simplifying, we have: y = 27x - 54.
To find the x-coordinate of point Q, we set y = 0 in the equation of PQ.
0 = 27x - 54
27x = 54
x = 54/27
x = 2.
Now, we have the x-coordinate of Q, which is x = 2.
To find the area enclosed between the curve, PQ, and the x-axis, we need to integrate the equation of the curve between the x-values of 0 and 2.
The equation of the curve y = x^3.
The area can be found by integrating y with respect to x between the limits 0 and 2.
∫[0 to 2] (x^3) dx = [(x^4)/4] [0 to 2] = [(2^4)/4] - [(0^4)/4] = 16/4 - 0 = 4 - 0 = 4 units^2.
Therefore, the correct area of the region enclosed between the curve, PQ, and the x-axis is indeed 4 units^2, not 27/4.