How do you find the equation of a tangent/normal line of a curve at a given point?
To find the equations of the tangent and the normal (perpendicular) lines at a given point on a curve, you need to know two things: 1.) the slope of the curve at the given point and 2.) a point (x, y) that each line passes through. Once we know this information, we can use it to determine the desired equation for each line.We can use the above slope and point information to find the desired equations by plugging this information into the “point-slope” form of the equation of a straight line, i.e., y ‒ y1 = m(x – x1), where m is the slope of the line and (x1, y1) is a specific point that the line passes through. We know that both the tangent and normal lines pass through and intersect at a given point (x1, y1) (a point of tangency) on the given curve. To find the slope of a curve and of the tangent line to the curve at a given point (x1, y1) on the curve, we need to know the equation, y = f(x), which algebraically represents the given curve and which defines a function f. Once we know y = f(x), we can then find the derivative f ′(x) of the function f. Graphically, the derivative f ′(x) is the slope of a curve at any point (x, y) on the curve, as well as the slope of the tangent line to the curve at the same point (x, y), provided f ′(x) exists; therefore, the slope of a given curve and of the tangent line to the curve at a given point (x1, y1) on the curve is given by m = f ′(x) = f '(x1), provided the derivative f ′(x) at x = x1 exists. Therefore, the equation of a tangent line to a curve at a given point (x1, y1) is found as follows:y ‒ y1 = m(x ‒x1)y ‒ y1 = [f '(x1)](x ‒ x1) y ‒ y1 + y1 = [f '(x1)](x ‒ x1) + y1 y + 0 = [f '(x1)](x ‒ x1) + y1 y = [f '(x1)](x ‒ x1) + y1 We know that the slopes of two perpendicular lines are negative reciprocals of one another, i.e., if m is the slope of one, then the slope of the other is equal to ‒1/m, i.e., their product, m(‒1/m), equals ‒1.Since the line normal to a curve at a given point (x1, y1) is also the same line that is perpendicular to the tangent line to the curve at the same point (x1, y1), then the slope of the normal line = ‒1/m = ‒1/f ′(x).Therefore, the equation of a normal (perpendicular) line to a curve at a given point (x1, y1) is found as follows: y ‒ y1 = m(x ‒x1)y ‒ y1 = [‒1/ f '(x1)](x ‒ x1)y ‒ y1 + y1 = [‒1/ f '(x1)](x ‒ x1) + y1y + 0 = [‒1/ f '(x1)](x ‒ x1) + y1y = [‒1/ f '(x1)](x ‒ x1) + y1
At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is...?
4x^2y - pi cos y = 5pi d/dx(4x^2y - pi cos y) = d/dx(5pi) 8xy + 4x^2(dy/dx) + pi siny(dy/dx) = 0 4x^2(dy/dx) + pi siny(dy/dx) = -8xy dy/dx(4x^2 + pi siny) = -8xy dy/dx = -8xy/(4x^2 + pi siny) At (1,pi), slope = -8(1)(pi)/(4(1^2) + pi sin(pi)) = -8pi/4 = -2pi Tangent line's equation is y - pi = -2pi(x - 1) y - pi = -2pix + 2pi y = -2pix + 3pi
What is the slope of the line tangent to the curve 3y^2 - 2x^2 = 6 - 2xy at the point (3,2)?
the slope of the tangent line to a point on the curve is the derivative of the function evaluated at that point, so: implicitly differentiate: 3y^2 - 2x^2 = 6 - 2xy take d/dx of each term: 6y dy/dx -4x dx/dx = -2xdy/dx -2y dx/dx noting that dx/dx =1, we get: (6y+2x)dy/dx=4x-2y dy/dx=(4x-2y)/(2x+6y) at (3,2): dy/dx=(12-4)/(6+12) = 1/2
At the given point, find the slope of the curve, the line that is tangent to the curve or the line that is? ?
3x^2y - pi cos y = 4pi, normal at (1,pi) => 3x^2*dy/dx+y*6x-pi(-sin y)*dy/dx = 0 => dy/dx = -(6 x y)/(3 x^2+pi sin(y)) normal at (1,pi) => slope = -(6 (1)(pi))/(3 (1)^2+pi sin(pi)) etc QED
At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that....?
Implicit differentiation gives 4y^3*(dy/dx) + 3x^2 = 2y*(dy/dx) + 10 Substitute x = 0, y = 1 and find dy/dx.
Find the Tangent of the Curve?
With x^6*y^6 = 64 = 2^6, we have xy = 2 and so y = 2/x and dy/dx = -2/x^2. At x = 2, the slope of the tangent is: dy/dx (evaluated at x = 2) = -2/2^2 = -1/2. Then, the tangent line has a slope of -1/2 and passes through (2, 1). This gives the equation of the line to be: y - 1 = (-1/2)(x - 2) ==> y = (-1/2)x + 2. I hope this helps!
What is the slope of the normal to the curve xy=6?
divide both side with x such thatxy/x=6xy = 6/xnow in order to get the slope of the function find the derivative d/dx of (6/x)d/dx(6/x) = -6/x^2 which is the slope
How do I find the tangent line to the curve y = e^(x/2) which goes through the point (0,0)?
HiI've used simple differentiation and substitution of x coordinates.Hope this helpedIf you have any queries please askThanks
What does mean to find slope of a line tangent to a graph?
Yes you are right that tangent line touches a curve at a single point . But It is a line so it also passes through infinite points and it does have a slope.Now lets say you are only given the point of contact from the curve. In that case how will you find the slope of tangent line?In that case you look for other information.For Example lets say you're given a circle with centre (h,k) and the point of contact of tangent is (x1,y1).How will you find the equation of tangent since you're only given point of contact ?We know that radius is perpandicular to tangengt at point of contact so the line segment joining (h,k) and (x1,y1) is perpandicualr to tangent at (x1,y1). Using this information you can find the slope of tangent. Since the slope of tangent will be the negative reciprocal of the slope of radius.Once you get the slope, deriving the equation is child's play.Now, slope of radius = (k-y1)/(h-x1)therefore, slope of tangent is = -(h-x1)/(k-y1)So the moral is that in such cases where only point of contact of tangent is given. You use other infomation you can have based on basic geometry and reasoning to move forward.
How do I determine the tangent line of the curve [math]xy^3+2y-2x=1[/math] at the point [math](x,y) = (1,1)[/math]?
The way I would do it is to first find the normal to the curve and get the tangent from that.Write [math]f(x,y) = x y^3 + 2 y - 2 x - 1[/math]. Find[math]\frac{\partial f}{\partial x} = y^3 - 2[/math][math]\frac{\partial f}{\partial y} = 3 x y^2 + 2[/math]At the point (1,1) these are -1 and 5. So a normal vector is[math]\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (-1,5).[/math]A tangent vector is this rotated by 90º so (5,1) which has gradient [math]m=\frac15[/math].A vector equation for the tangent line is [math](x,y) = (1,1) + t (5,1)[/math].To get the y=m x + c version, we know the [math]m=\frac15[/math]. Substitute the known point [math]1 = \frac15 \times 1 + c[/math], so [math]c = \frac45[/math] and the equation is [math]y = \frac15 x + \frac45.[/math]The advantage of this method is that you don’t need to solve any equations, if you have a curve in implicit form and a known point you can always find the partial derivatives and get the normal. It can also be extended to surfaces in 3D.