1. Given the curve
a. Find an expression for the slope of the curve at any point (x, y) on the curve.
b. Write an equation for the line tangent to the curve at the point (2, 1)
c. Find the coordinates of all other points on this curve with slope equal to the slope at (2, 1)
sorry here's the equation:
x+xy+2y^2=6
There was a mistake in the calculation of dx/dy. It should have been expressed in terms of y and not x.
Given : x+xy+2y^2=6
(a)
x(1+y)=6-2y²
x=(6-2y²)/(1+y)
differentiate with respect to y to get
dx/dy=-4y/(y+1)-(6-2y^2)/(y+1)^2
So
f'(y)=dy/dx
= 1/(dx/dy)
= -(1/2)(y^2+2y+1)/(y^2+2y+3)
(b) For the point (2,1), first check that it lies on the curve.
f'(y)=
f'(1)=-(1/2)(1^2+2*1+1)/(1^2+2*1+3)
=-1/3
Tangent line with slope -1/3 passing through (2,1) is:
(y-1) = (-1/3)(x-2)
(c)
Now solve for
f'(y)=-1/3
to get x=-3 or x=1
Please verify all the numerical values.
a. Well, the expression for the slope of a curve at any point (x, y) is the derivative of the function that defines the curve. So, in this case, we need to find the derivative of the curve equation.
b. To find the equation of the line tangent to the curve at the point (2, 1), we'll use the point-slope form of a line equation. This form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. So, we just need to plug in the coordinates of the point (2, 1) and the slope we found in part (a).
c. To find the coordinates of all other points on the curve with a slope equal to the slope at (2, 1), we'll need to find the points where the derivative of the curve equation is equal to the slope we found in part (a). This will give us the x-coordinates of the points. We can then substitute these x-coordinates back into the curve equation to find the corresponding y-coordinates.
To answer these questions, we need to find the derivative of the curve and use it to find the slope at any point.
a. Finding the expression for the slope of the curve at any point (x, y) on the curve requires finding the derivative of the curve. The derivative measures the rate of change of the function with respect to x. Let's denote the curve as y = f(x).
To find the derivative, we can use differentiation rules. Differentiate the function f(x) with respect to x. The slope of the curve at any point (x, y) on the curve will be given by:
slope = dy/dx = f'(x)
b. To find the equation for the line tangent to the curve at the point (2, 1), we need the slope at that point. Let's denote it as m. So, we need to find m and then use the point-slope form of a line equation.
1. Find the slope at (2, 1):
Substitute x = 2 into the expression for the derivative obtained in part a:
slope = f'(2)
2. Use the point-slope form of a line equation:
The equation of the tangent line is given by:
y - y1 = m(x - x1)
Substitute m (slope at (2, 1)) into the equation:
y - 1 = slope(x - 2)
c. To find the coordinates of all other points on the curve with slope equal to the slope at (2, 1), we need to set the derivative of the curve equal to the slope at (2, 1) and solve it for x.
Set f'(x) = slope (slope at (2, 1)).
Solve the equation for x to find the x-coordinates of the other points.
Once you have the x-coordinates, substitute them back into the equation of the curve to find the corresponding y-coordinates.
x+xy+2y^2=6
(a)
x(1+y)=6-2y²
x=(6-2y²)/(1+y)
differentiate with respect to y to get
dx/dy=-4x/(x+1)-(6-2x^2)/(x+1)^2
So
dy/dx
= 1/(dx/dy)
= -(1/2)(x^2+2x+1)/(x^2+2x+3)
(b) For the point (2,1), first check that it lies on the curve.
f'(2)=-(1/2)(2^2+2*2+1)/(2^2+2*2+3)
=-9/22
(c)
Now solve for
f'(x)=-9/22
to get x=-4 or x=2
Verify all the numerical values.