Quizzes & Puzzles0 min ago
Maths integration help needed if possible
1 Answers
hey
im really stuck on a integration question for my A level maths course.
heres the question, and hope someone can help me out.
4. The curve y=12 - 3x^2 has a tangent at the point p (2,0) on the curve.
(a) find Integral 0,2 (12-x^2)dx
(b) (i) find the gradient of the curve at point P
(ii) Point Q is where the tangent crosses the y axis, find the coordinates of this
(c) find the area of the region enclosed between the tangent, curve, and y axis.
thanks for ANY help
its much much apprechiated.
im really stuck on a integration question for my A level maths course.
heres the question, and hope someone can help me out.
4. The curve y=12 - 3x^2 has a tangent at the point p (2,0) on the curve.
(a) find Integral 0,2 (12-x^2)dx
(b) (i) find the gradient of the curve at point P
(ii) Point Q is where the tangent crosses the y axis, find the coordinates of this
(c) find the area of the region enclosed between the tangent, curve, and y axis.
thanks for ANY help
its much much apprechiated.
Answers
Best Answer
No best answer has yet been selected by acroviak. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.(a) I assume you mean the integral of (12 - 3x^2)dx?
If so then it is [12x - x^3] from x=0 to x=2, ie (12*2-2^3) - (12*0-0^3) = 24-8=16 square units.
(b)(i) The gradient is given by the value of dy/dx at point P.
dy/dx = -6x. So if x=2, the gradient is -6*2=-12.
(ii) Assume the equation of the tangent is y = -12x + a. Since it goes through the point (2,0) then 0=-24+a, ie a = 24
So the equation of the tangent is y = 24-12x.
That line cuts the y axis when x=0, ie when y=24, ie the coordinates of Q are (0,24).
(c) The area enclosed by the tangent and the x and y axes is a triangle of size 0.5*24*2=24 square units. From part (a) the area under the curve is 16 square units. So the area of the region between the tangent, curve and y axis is 24-16=8 square units.
If so then it is [12x - x^3] from x=0 to x=2, ie (12*2-2^3) - (12*0-0^3) = 24-8=16 square units.
(b)(i) The gradient is given by the value of dy/dx at point P.
dy/dx = -6x. So if x=2, the gradient is -6*2=-12.
(ii) Assume the equation of the tangent is y = -12x + a. Since it goes through the point (2,0) then 0=-24+a, ie a = 24
So the equation of the tangent is y = 24-12x.
That line cuts the y axis when x=0, ie when y=24, ie the coordinates of Q are (0,24).
(c) The area enclosed by the tangent and the x and y axes is a triangle of size 0.5*24*2=24 square units. From part (a) the area under the curve is 16 square units. So the area of the region between the tangent, curve and y axis is 24-16=8 square units.
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