find the values of p and q if slope of the tangent to the curve xy + pq + qy = 2(1, 1) is 2
Good answer.
To find the values of p and q, we need to use the equation of the curve and the given information about the slope of the tangent line.
Step 1: Start with the equation of the curve
The equation of the curve is given as xy + pq + qy = 2.
Step 2: Differentiate the equation with respect to x
Differentiating the equation with respect to x will give us the derivative of y with respect to x, which represents the slope of the tangent line.
Differentiating xy + pq + qy = 2 with respect to x gives us:
y + xy' + qy' = 0.
Step 3: Simplify the differentiated equation
Combine the terms with y' to get:
y' (x + q) = -y.
Step 4: Calculate the slope of the tangent line at the point (1, 1)
Substitute x = 1 and y = 1 into the simplified equation from step 3:
1' (1 + q) = -1.
Simplifying further:
1 + q = -1.
Step 5: Solve for q
Subtract 1 from both sides of the equation:
q = -1 - 1.
q = -2.
Step 6: Solve for p
Substitute the value of q into the original equation:
xy + pq + qy = 2.
(1)(1) + p(-2) + (-2)(1) = 2.
1 - 2p - 2 = 2.
-2p - 1 = 2 - 1.
-2p - 1 = 1.
-2p = 1 + 1.
-2p = 2.
p = 2 / -2.
p = -1.
Therefore, the values of p and q are p = -1 and q = -2.
To find the values of p and q, we need to find the equation of the tangent line to the curve at the point (1, 1) and then determine the slope of that tangent line.
First, let's find the equation of the curve. We have the equation xy + pq + qy = 2.
To find the equation of the tangent line at the point (1, 1), we need to find the derivative of the curve with respect to x and y separately, and then use the point-slope form of the equation of a line.
Differentiating the equation with respect to x:
d/dx(xy) + d/dx(pq) + d/dx(qy) = d/dx(2)
y + x(dy/dx) + 0 + q(dy/dx) + qy(dx/dx) = 0
(dy/dx)(x + q) + y + qy = 0
Differentiating the equation with respect to y:
d/dy(xy) + d/dy(pq) + d/dy(qy) = d/dy(2)
x + 0 + q(dy/dy) + qy = 0
x + q + qy = 0
Now we need to solve this system of equations for dy/dx (slope):
(dy/dx)(x + q) + y + qy = 0 ---(1)
x + q + qy = 0 ---(2)
Let's substitute x = 1 and y = 1 in equations (1) and (2):
(dy/dx)(1 + q) + 1 + q(1) = 0 ---(3)
1 + q + q(1) = 0 ---(4)
Equation (3) simplifies to:
(dy/dx)(1 + q) + 1 + q = 0
Rearranging equation (3), we get:
(dy/dx)(1 + q) = -1 - q
Now, we can substitute q = 2 into equation (4):
1 + (2) + (2)(1) = 0
1 + 2 + 2 = 0
5 = 0
Since we have reached a contradiction, it means that there are no real values of p and q for which the slope of the tangent to the curve at the point (1, 1) is 2.
Hence, there are no values of p and q that satisfy the given conditions.
xy' + y + qy' = 0
y' = -y/(x+q)
at (1,1) y' = -1/(1+q) = 2
so, q = -3/2
so, at (1,1)
1 - 3/2 p - 3/2 = 2
p = -5/3
xy + 5/2 - 3y/2 = 2
and the tangent line is
y-1 = 2(x-1)
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plot xy + 5/2 - 3y/2 = 2, y-1 = 2(x-1), x=0..2