Use implicit differentiation to find an equation of the tangent line to the curve at the given point. 4x2 + xy?
4x² + xy + 4y² = 9 Differentiate implicitly w.r.t. x: 8x + y + x dy/dx + 8y dy/dx = 0 Collect dy/dx terms on the left and move everything else to the right: x dy/dx + 8y dy/dx = -8x - y => (x + 8y) dy/dx = -(8x + y) => dy/dx = - (8x + y) / (x + 8y) At the point (1, 1) we get dy/dx|(1, 1) = -(9)/9 = -1. So the equation of the tangent line is (y - 1) = (-1) (x - 1) i.e. y = 2 - x.
Use implicit differentiation to find an equation of the tangent line to the curve at the given point?
6x^2 + xy + 6y^2 = 13 12x + y + xy' + 12yy' = 0 y'.(x + 12y) = - (12x + y) So, y' = - (12x + y) / (x + 12y) is the slope of equations of tangent lines to the curve at any point. ... Hope this helps!
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.?
x² + y² = (2x² + 2y² - x)² Differentiating, you have 2x + 2y (dy/dx) = 2 (2x² + 2y² - x) (4x + 4y (dy/dx) - 1) x + y (dy/dx) = (2x² + 2y² - x) (4x + 4y (dy/dx) - 1) x + y (dy/dx) = (4x - 1) (2x² + 2y² - x) + 4y (dy/dx) (2x² + 2y² - x) x - (4x - 1) (2x² + 2y² - x) = 4y (dy/dx) (2x² + 2y² - x) - y (dy/dx) x - (4x - 1) (2x² + 2y² - x) = dy/dx (4y (2x² + 2y² - x) - y) dy/dx = [x - (4x - 1) (2x² + 2y² - x)] / [4y (2x² + 2y² - x) - y] At the point (0, 1/2), dy/dx = [0 - (0 - 1) (0 + 2(1/2)² - 0)] / [4(1/2) (0 + 2(1/2)² - 0) - (1/2)] dy/dx = [-(-1)(1/2)] / [4(1/2) (1/2) - (1/2)] dy/dx = [1/2] / [1 - (1/2)] dy/dx = [1/2] / [1/2] dy/dx = 1 The equation of the tangent line is y - 1/2 = 1 (x - 0) y = x + 1/2 Here's a picture: http://www.wolframalpha.com/input/?i=x%C...
Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=12 at the pt (6,1)
implicit differentiation of xy^3 + xy = 12 uses the product rule: 1y^3 + x*3y^2 dy/dx + 1y + x*1 dy/dx = 0 y^3 + 3xy^2 dy/dx + y + x dy/dx = o keeping the dy/dx terms on left and moving the y^3 and y to other side 3xy^2 dy/dx + x dy/dx = -y^3 -y factor out dy/dx dy/dx( 3xy^2 + x) = -y^3 - y then dy/dx = (-y^3 - y)/(3xy^2+ x) putting in the point (6,1) dy/dx = (-1 -1)/(18+6) = -2/24 = -1/12 thus, m = -1/12 your line equation is now y = -1/12 x + b putting in (6,1) again, 1 = (-1/12)(6) +b 1+1/2 = b 3/2 = b y = -1/12x + 3/2
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.?
... 7x² + xy + 7y² = 15 ∴ diff... w.r.t. x, by Chain Rule, ... 7(2x) + ( x.y' + y.1 ) + 7(2y).y' = 0 ∴ (x+14y).y' = -14x-y ∴ y' = -(14x+y) / (x+14y) ∴ slope m = [ y' at (1,1) ] = - (14+1) /(1+14) = -1 ∴ the req'd eq of tgt is ... y - 1 = -1( x - 1 ) = 0 ∴ y - 1 = -x + 1 ∴ x + y = 2 ........................................... Ans. _______________________
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.?
y sin 16x = x cos 2y 16y cos(16x) + y' sin(16x) = cos(2y) - 2xy' sin(2y) To find the slope m, replace y' with m, x with π/2, and y with π/4. 16*π/4 cos(16*π/2) + m*sin(16*π/2) = cos(2*π/4) - 2*π/2*m*sin(2*π/4) 4π cos(8π) + m*sin(8π) = cos(π/2) - π*m*sin(π/2) 4π *1 + m*0 = 0 - π*m*1 4π = -π*m m = -4 Equation of the tangent line: y - π/4 = -4(x - π/2) y = -4x + 2π + π/4 y = -4x + 9π/4
Use implicit differentiation to find an equation of the tangent line to the ellipse at point (1,1)?
8x² + xy + 8y² = 17 16x + x dy/dx + y + 16y dy/dx = 0 16x + y = dy/dx (-16y - x) dy/dx = (16x + y) / (-16y - x) at (1,1) dy/dx = (16 + 1) / (-16 - 1) = -1 y = mx + b 1 = (-1)(1) + b b = 2 Answer: y = -x + 2
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.?
First use implicit differentiation to find the slope of the tangent line at the given point. d/dx (9x²) = 18x. Using the product rule: d/dx (xy) = y + x * dy/dx. Using the chain rule: d/dx (9y²) = d/dy (9y²) * dy/dx = 18y * dy/dx. And of course d/dx (19) = 0. Hence 18x + y + x * dy/dx + 18y * dy/dx = 0. ∴ (x + 18y) dy/dx = -(18x + y). ∴ dy/dx = -(18x + y)/(x + 18y). Hence at x = 1, y = 1, dy/dx = -19/19 = -1. So the equation of the tangent line at (1, 1) is: y - 1 = -(x - 1) = 1 - x. Or y = 2 - x
Calc...use implicit differentiation to find an equation of the tangent line to the curve at the given point?
The equation to the tangent at the desired point is: y = x / [3^(1/2)] + 4. This is found as follows. Write the original equation of the curve as: x^(2/3) + y^(2/3) = 4. (### --- see below.) Therefore, (2/3) x^(-1/3) + (2/3) y^(-1/3) dy/dx = 0. Thus: dy/dx = - (x/y)^(-1/3) = - (y/x)^(1/3). Then the equation of the tangent line at the point (-3*3^(1/2), 1) --- it IS on the curve, by the way! --- is: (y - 1) / [x - {-3*3^(1/2)}] = {the VALUE of dy/dx at (-3*3^1/2^, 1)}, which is {- 1 / [- 3*3^(1/2)]^(1/3)} = + 1/[3^(1/2)]. Therefore (y - 1) / [x - {-3*3^(1/2)}] = + 1 / [3^(1/2)], or y - 1 = x / [3^(1/2)] + 3, i.e. y = x / [3^(1/2)] + 4. [Final result.] QED CHECK: This clearly has the right SLOPE, + 1 / [3^(1/2)]; so let's just check that the point (-3*3^(1/2), 1) IS on it: At the given point, y = 1; and x / [3^(1/2)] + 4 = - 3*3^(1/2) / [3^(1/2)] + 4 = - 3 = 4 = 1. PHEW! It checks out. Live long and prosper. ### P.S. PLEASE write things like this equation for the astroid, or indeed anything else containing fractional powers or exponents, as I did : x^(2/3) + y^(2/3) = 4, rather than as you did, x^2/3^ + y^2/3^ = 4. The second ' ^ ' in each of your own terms on the LHS is confusing. The symbol ' ^ ' means "what immediately follows is to be read as an exponent of what preceded it." It's only used ONCE to indicate that exponent or power. If the exponent is more than just a number or a single symbol, the PARENTHESES around the expression for the exponent clearly delimit it. (You seem to have used the ' ^ ' symbol to signal that you are first beginning, and then that you are finally ending, the exponent. But that's confusing because it's NOT the usual convention. Conventionally, it would be interpreted as you intending to divide by 3 to some power, but then neglected to type it!)