A line perpendicular to another line or to a tangent line is o...

Question

A line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point P.
y= √x; P(4, 2)

y= √x; P(4, 2)

Accepted Answer

Welcome back, everyone. Determine the equation of the line perpendicular to the tangent line of the curve Y equals X to the power of three halves at the point P 9.25. We're given 4 answer choices A Y equals 29 x plus 25. B-29 X plus 29 C 29 X plus 29 and D 29 X plus 25. So let's understand what we're trying to achieve in this problem. Let's suppose we're given a curve Y equals F of X. Let's say we have a point of interest which coordinates X, not Y not. And let's suppose that we have a tangent line passing through that point. In this context, we want to find a normal line, which is the line that is perpendicular to the tangent line at the same point, right? So, Basically, our X. Y knot is equal to 9.27. That's the point of tangency point P, right? So what we can do first of all is identify the slope of the tangent line because if we know the slope of the tangent line, we will be able to find the slope of the normal line. And we have to recall that for this purpose we need to find the derivative of the function. So let's identify the derivative of X to the power of three halves, and using the power rule, this is 3 halves x to the power of 3/hals minus 1. So that's x to the power of 1/2, and we can rewrite this as 3 halves square root of X. And now the slope is simply the derivative of Y at the given point, so we want to evaluate Y at 0.9. So we're going to substitute 9 and we're going to get 3 halves multiplied by square root of 9, which is 9 halves. And this is M1. We're going to we're going to call it M1, which represents the slope of the tangent line passing through the point P. So essentially we're going to say that M1 is equal to 9 halves. The second step, well this is where we can identify the slope of the normal line M2, and we know that the product of two lines that are perpendicular to each other is -1. So M2, the slope. Of the normal line is simply -1 divided by m1. Substituting M1, we get -1 divided by nine halves. We're taking the reciprocal value and we get negative 2 9s. So it's either B or D, right? Because we know that slope is -29s. We can cross out A and C. Those have positive slopes, we need the negative slope. And step number 3, well, we can apply the point slope formula. Y minus Y1, so Y minus Y1? Essentially, we can take the given point because this is the common point for the two lines. So Y minus 27. Should be equal to the slope in this case -29s multiplied by x minus X1. We can take the same point. It has a coordinate X knot of 9, so we're subtracting 9. And all that we have to do is simply solve for Y. So Y equals -2 9 X. Then distributing the terms we get minus 2 9s multiplied by -9. This gives us 2, and then we're adding 27, right? So Y equals -2 9 X plus 29, which corresponds to the inter choice B, and that's our final answer. Thank you for watching.

A line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point P.
y= √x; P(4, 2)

Key Concepts

Tangent Line

Normal Line

Finding the Derivative

Watch next

A line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point P.y= √x; P(4, 2)

Key Concepts

Tangent Line

Normal Line

Finding the Derivative

Watch next

A line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point P.
y= √x; P(4, 2)