Equation of Normal to Curve at (1,5): Solved?

In summary, the conversation was about finding the equation of a normal at a given point on a curve with a specific rule. The correct solution involved finding the slope of the tangent line and using it to derive the equation of the normal. The incorrect solution involved substituting values into the wrong equation.
  • #1
alexandravo
2
0
Really confused bout a question and finding the equation.

A normal is drawn at the point (1,5) on the curve defined by the rule y=x2+4. Find the equation of the normal.

I substituted the values x=1 and y=5 into the derived equation and got my answer to be x+2y=10? Is that correct?
 
Physics news on Phys.org
  • #2
alexandravo said:
Really confused bout a question and finding the equation.

A normal is drawn at the point (1,5) on the curve defined by the rule y=x2+4. Find the equation of the normal.

I substituted the values x=1 and y=5 into the derived equation and got my answer to be x+2y=10? Is that correct?
You got the slope right. But the point $(1,5)$ does not lie on the line $x+2y = 10$, so you need to adjust the constant $10$.
 
  • #3
If y=x^2+ 4, y'= 2x. (1, 5) is on the curve so the tangent line through (1, 5) has slope 2(1)= 2.
The tangent line is y(x)= 2x+ C where C is such that y(1)= 2(1)+ C= 5. C= 5- 2= 3. The tangent line is y= 2x+ 3.

You say "I substituted the values x=1 and y=5 into the derived equation". I assume you mean the equation y'= 2x. Yes, with x= 1, y', the slope, is 2 but there is no reason to calculate 2 times 5! y is 5, not x! And y' is the slope of the tangent line, not the constant term.
 
  • #4
Country Boy said:
If y=x^2+ 4, y'= 2x. (1, 5) is on the curve so the tangent line through (1, 5) has slope 2(1)= 2.
The tangent line is y(x)= 2x+ C where C is such that y(1)= 2(1)+ C= 5. C= 5- 2= 3. The tangent line is y= 2x+ 3.

You say "I substituted the values x=1 and y=5 into the derived equation". I assume you mean the equation y'= 2x. Yes, with x= 1, y', the slope, is 2 but there is no reason to calculate 2 times 5! y is 5, not x! And y' is the slope of the tangent line, not the constant term.
But we're looking for the normal, not the tangent!
 
  • #5
Opalg said:
You got the slope right. But the point $(1,5)$ does not lie on the line $x+2y = 10$, so you need to adjust the constant $10$.
Country Boy Is correct in stating that the slope of the tangent line is 2. So the tangent line equation is
y = 2x + 3
and the equation of the normal is then
y = -x/2 + 11/2
 

1. What is the equation of the normal to a curve at the point (1,5)?

The equation of the normal to a curve at a given point is the line that is perpendicular to the tangent line at that point. To find the equation, we first need to find the slope of the tangent line, which can be done by taking the derivative of the curve at the given point. Then, we can use the point-slope formula to find the equation of the normal.

2. How do you find the slope of the tangent line at a given point?

The slope of the tangent line at a given point can be found by taking the derivative of the curve at that point. This can be done using the power rule, product rule, quotient rule, or chain rule, depending on the complexity of the curve. Once the derivative is found, evaluate it at the given point to get the slope of the tangent line.

3. Can the equation of the normal to a curve change at different points?

Yes, the equation of the normal can change at different points on a curve. This is because the slope of the tangent line, and therefore the slope of the normal, can vary at different points on the curve. This is why we need to specify the point at which we want to find the equation of the normal.

4. How can the equation of the normal be used in real-world applications?

The equation of the normal can be used in various real-world applications, such as in physics, engineering, and economics. For example, in physics, the normal force acting on an object is equal to the weight of the object multiplied by the cosine of the angle between the object and the surface it is resting on. This can be represented by an equation using the normal vector.

5. Can the equation of the normal be used to find the equation of the curve?

No, the equation of the normal cannot be used to find the equation of the curve. The equation of the normal only gives us information about the slope and position of the normal line at a given point on the curve. To find the equation of the curve, we need to know more points on the curve or have additional information, such as the degree or type of the curve.

Similar threads

B Confusion on Implicit Differentiation
    • Calculus
Replies
16
Views
2K
I How to evaluate the enclosed area of this implicit curve?
    • Calculus
Replies
2
Views
2K
B Calculating shaded area: I'm getting discrepancies between methods
    • Calculus
Replies
3
Views
339
B Why is this definite integral a single number?
    • Calculus
Replies
20
Views
2K
The relation between the normal and the slope of a cylindrical curve
    • Calculus and Beyond Homework Help
Replies
2
Views
658
I Question about gradient, tangent plane and normal line
    • Calculus
Replies
1
Views
1K
Calculus Help locating a Math Equation to fit this curve
    • Science and Math Textbooks
Replies
10
Views
902
MHB How Do You Calculate the Length of the Curve from $x=\ln{4}$ to $x=\ln{5}$?
    • Calculus
Replies
10
Views
2K
I Gradient Vectors: Perpendicular to Level Curves
    • Calculus
Replies
4
Views
2K
MHB Finding a parameter for which a line is orthogonal to a curve
    • Calculus
Replies
1
Views
3K

Hot Threads

Recent Insights

Back
Top