Find the slope of the tangent line to the curve (a lemniscate)
2(x^2+y^2)^2=25(x^2–y^2) at the point
(–3,1) .
I know that we are supposed to use chain rule for this problem but i think i am messing up on the algebra. Can someone please help me, i keep on getting m= (42x^3-42y^2x)/(58x^2+58y^2)
2(x^2+y^2)^2=25(x^2–y^2)
2(x^2+y^2)^2-25(x^2–y^2)=0
Expand
2x^4+4x^2y^2+2y^4-25x^2+25y^2=0
Differentiate (implicitly) with respect to x, denote dy/dx as y'
8x^3+8xy^2+4x^2(2y)y'+8y^3y'-50x+50yy'=0
Collect terms and solve for y' in terms of x and y:
y'=(50x-8x^3-8xy^2)/(8x^2y+8y^3+50y)
At (-3,1), x=-3, y=1
y'=(-150+216+24)/(72+8+50)=9/13
Check for arithmetic errors.
I get
4(x^2 + y^2)(2x + 2yy') = 50x - 50y y'
.....
....
y' = ((50x - 8x^3 - 8xy^2)/[ (8x^2)y + 8y^3 - 50y)
plug in your point (-3,1) to get the slope
Take it from there.
Oh my goodness, thank you soooo much for writing it out i totally wasn't getting it! THank you so much :)
Do you think you coul check my arithmatic for another problem? It's
Use implicit differentiation to find the slope of the tangent line to the curve
y/x+6y=x^2–6 at the point (1–5/31)
I got [(x+6y)(y')-(y)(1+6y')]/(x+6y)^2=2x then y'=(2x+x^2+12yx+6y^2-y)/(x+6y+6). Is that right?
y/(x+6y) or (y/x)+6y?
As it is, the expression means the latter.
Since I see (x+6y)y' somewhere, I think you mean the first interpretation.
y/(x+6y)=x^2–6
Cross multiply to get
y=(x^2-6)(x+6y)
y'=2x(x+6y)+(x^2-6)(1+6y')
I get
y'=-(12xy+3x^2-6)/(6x^2-37)
Check my arithmetic and differentiation and take it from here.
To find the slope of the tangent line to the curve, we need to differentiate the given equation and then evaluate it at the given point. Let's break down the process step by step.
Step 1: Differentiate the equation using the chain rule.
Start by differentiating both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating y, we need to apply the chain rule.
2(x^2+y^2)^2 = 25(x^2–y^2)
Differentiating both sides:
d/dx[2(x^2+y^2)^2] = d/dx[25(x^2–y^2)]
Using the chain rule, we get:
4(x^2+y^2)(2x + 2yy') = 50x - 50yy'
Step 2: Find the derivative dy/dx.
Rearranging the equation, we have:
8x(x^2+y^2) + 8y^3y' = 50x - 50yy'
Now, solve for dy/dx by isolating y':
8y^3y' + 50yy' = 50x - 8x(x^2+y^2)
Factor out y' from the left side:
y'(8y^3 + 50y) = 50x - 8x(x^2+y^2)
Divide both sides by (8y^3 + 50y):
y' = (50x - 8x(x^2+y^2))/(8y^3 + 50y)
This is the expression for the derivative dy/dx, or the slope of the tangent line.
Step 3: Plug in the point (-3,1).
To find the slope of the tangent line at the given point (-3,1), substitute x = -3 and y = 1 into the derived expression for dy/dx:
y' = (50(-3) - 8(-3)((-3)^2+1^2))/(8(1)^3 + 50(1))
Simplifying this expression:
y' = (-150 + 72)/(8 + 50)
= -78/58
= -39/29
Therefore, the slope of the tangent line to the curve at the point (-3,1) is -39/29.
Please double-check your algebra to ensure correctness.