Find the equation of the tangent line to the curve (piriform)
y^2=x^3(4−x)
at the point (2,16−−�ã).
a. Find dy/dx at x=2.
dy/dx=
b. Write the equation of the tangent line to the curve.
y^2 = 4x^3 - x^4
2yy' = 12x^2 - 4x^3
y' = (6x^2 - 2x^3)/y
y(2) = 4
y'(2) = (6*4-2*8)/4 = 2
So, now you have a point and a slope. The tangent line at (2,4) is
y-4 = 2(x-2)
2y dy/dx = x^3 (-1) + 3x^2 (4-x)
dy/dx = (-x^3 + 12x^2 - 3x^3)/(2y)
= (6x^2 - 2x^3)/y
I can't make out your point (2, 16−−�ã)
but it should be easy for you
just plug in x=2 and y = whatever into the dy/dx
and that becomes your slope
then use the grade 9 way you learned to find the equation of a line with a given slope and a given point.
To find the equation of the tangent line to the curve at the point (2, 16):
a. Find dy/dx at x=2.
To find dy/dx, we need to differentiate both sides of the equation y^2 = x^3(4 - x) with respect to x.
On the left-hand side, we can rewrite y^2 as y * y, so:
2y * dy/dx = d/dx (x^3(4 - x))
Using the product rule for differentiation, we have:
2y * dy/dx = 3x^2(4 - x) + x^3(-1)
Simplifying further:
2y * dy/dx = 12x^2 - 3x^3 - x^3
Substituting the values x = 2 and y = 16, we get:
2(16) * dy/dx = 12(2)^2 - 3(2)^3 - (2)^3
32 * dy/dx = 48 - 24 - 8
32 * dy/dx = 16
Dividing both sides by 32:
dy/dx = 16/32
dy/dx = 1/2
b. Write the equation of the tangent line to the curve.
We can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) represents the point (2, 16) and m represents the slope dy/dx = 1/2.
Plugging in the values, we have:
y - 16 = (1/2)(x - 2)
Simplifying:
y - 16 = (1/2)x - 1
Rearranging the equation:
y = (1/2)x + 15
Therefore, the equation of the tangent line to the curve is y = (1/2)x + 15.