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Chapter 10: Problem 36

The tangent and the normal at a point \(\mathrm{P}\left(t, \mathrm{e}^{-t^{2} / 2}\right)\) on the curve \(y=\mathrm{e}^{-x^{2} / 2}\) meet the \(x\)-axis in \(\mathrm{T}\) and \(\mathrm{G}\) respectively, and \(\mathrm{N}\) is the foot of the ordinate from \(\mathrm{P}\). Show that \(\mathrm{G}=\left[t\left(1-\mathrm{e}^{-t^{2}}\right), 0\right]\) and if \(\mathrm{NT} . \mathrm{GN}=\mathrm{e}^{-1}, \quad\) find the length of \(\mathrm{PN}\). \((\mathrm{AEB})^{\prime} 75\)

Short Answer

Expert verified
PN = e^{-1/2}.

Step by step solution

01

Define the point P and the curve

Point P is given as \(P = (t, e^{-t^{2} / 2})\) on the curve \(y = e^{-x^{2} / 2}\).
02

Calculate the derivative of the curve

The curve is \(y = e^{-x^{2} / 2}\). Differentiate with respect to \(x\) to find the slope: \[\frac{dy}{dx} = -x e^{-x^{2} / 2}\]
03

Find the slope at point P

At point P, where \(x = t\), the slope is \[ -t e^{-t^{2} / 2}\]
04

Write the equation of the tangent line

The equation of the tangent line at P is \(y - e^{-t^{2}/ 2} = -t e^{-t^{2}/ 2}(x - t)\). Since it meets the x-axis at T, set \(y = 0\) and solve for \(x\): \[0 - e^{-t^{2}/ 2} = -t e^{-t^{2}/ 2}(x - t)\]Simplifying we get \[x = t + e^{t^{2}}\]Thus, \(T = (t + e^{t^{2}}, 0)\)
05

Write the equation of the normal line

The slope of the normal is the negative reciprocal of the slope of the tangent: \[\frac{1}{t e^{-t^2/2}}\] Thus, the equation of the normal at P is \[y - e^{-t^{2}/ 2} = \frac{1}{t e^{-t^{2}/ 2}} (x - t)\]Setting \(y = 0\) to find the x-intercept where G meets the x-axis: \[0 - e^{-t^{2}/ 2} = \frac{1}{t e^{-t^{2}/ 2}} (x - t)\]Simplifying we get \[x = t - t e^{t^{2}}\]Thus, \(G = (t(1 - e^{t^{2}}), 0)\) which simplifies to \[G = \[t \(1 - e^{-t^{2}}\), 0\] \]
06

Find the coordinates of N

The point N is the foot of the perpendicular from P to the x-axis, which means its coordinates are \(N = (t, 0)\).
07

Calculate the distances NT and GN

Distance NT: \[(t + e^{t^{2}}) - t = e^{t^{2}}\]Distance GN: \[t - t(1 - e^{-t^{2}})= t e^{-t^{2}}\]
08

Use the given condition to find PN

Given that \(NT \cdot GN = e^{-1}\), substitute values: \[e^{t^{2}} \cdot t e^{-t^{2}} = e^{-1}\] Simplifying: \[t = 1\]The length of PN is simply \[e^{-t^{2}/2}\When \(t=1\) PN = e^{-1/2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
To understand the concepts of tangent and normal lines, it's essential to grasp the foundation of derivatives in calculus. A derivative represents the rate at which a function is changing at any given point. In simpler terms, it's the slope of the curve or the steepness of the function at that point.

For our example, the curve is defined by the equation: \( y = e^{-x^{2}/2} \). To find the derivative, we use the chain rule to differentiate with respect to \( x \):
\[ \frac{dy}{dx} = -x e^{-x^{2}/2} \]
This expression gives us the slope of the curve at any point \( x \). When evaluated at a specific point, it tells us how steep the curve is at that spot.
Equation of Tangent Line
The tangent line to a curve at a point is the straight line that just 'touches' the curve at that point. Its slope is the same as that of the curve at the same point. Using point \( P = (t, e^{-t^{2}/2}) \), we find the slope from the derivative. The slope at point \( P \) would then be:
\[ -t e^{-t^{2} / 2} \]
To write the equation of the tangent line, we use the point-slope form of a line equation:
\[ y - e^{-t^{2}/2} = -t e^{-t^{2}/ 2}(x - t) \]
This equation can be simplified to find where the tangent meets the x-axis by setting \( y = 0 \) and solving for \( x \). This intersection point is denoted as \( T \). By solving, we get:
\[ T = (t + e^{t^{2}}, 0) \]
Equation of Normal Line
The normal line to a curve at a given point is perpendicular to the tangent line at that point. To find the equation of the normal line, we first find the slope of the normal, which is the negative reciprocal of the slope of the tangent. For our curve, the slope of the normal line at point \( P \) is given by:
\[ \frac{1}{t e^{-t^2 / 2}} \]
Using point-slope form, the equation for the normal at \( P \) is:
\[ y - e^{-t^{2}/ 2} = \frac{1}{t e^{-t^{2}/ 2}} (x - t) \]
To find the intersection with the x-axis, set \( y = 0 \) and solve for \( x \). This gives us the point \( G \):
\[ G = \bigg[ t \big(1 - e^{-t^{2}}\big), 0 \bigg] \]
Distance Calculation
To solve the exercise, we need to find the distances between certain points. Let's start with point \( N \), the foot of the ordinate from \( P \), which is directly below \( P \) on the x-axis. This point is:
\[ N = (t, 0) \]
Next, we need to compute the distances:
  • Distance \( NT \): \( NT = (t + e^{t^2}) - t = e^{t^2} \)
  • Distance \( GN \): \( GN = t - t(1 - e^{-t^2}) = t e^{-t^2} \)

Given the condition \( NT \cdot GN = e^{-1} \), we substitute the values:
\[ e^{t^2} \times t e^{-t^2} = e^{-1} \]
Simplifying this condition, we find:
\[ t = 1 \]
Finally, to find the length of \( PN \), which is the distance between \( P \) and \( N \), we use the y-coordinate of point \( P \) when \( t = 1 \):
\[ PN = e^{-1/2} \]

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Most popular questions from this chapter

A point \(\mathrm{P}\) moves so that it is equidistant from \(\mathrm{A}\) and \(\mathrm{B}\). The locus of the set of points \(\mathrm{P}\) is: (a) a circle on \(\mathrm{AB}\) as diameter, (b) a line parallel to \(\mathrm{AB}\), (c) the perpendicular bisector of \(\mathrm{AB}\), (d) a parabola with focus \(\mathrm{A}\) and directrix \(\mathrm{B}\), (e) none of these.

Find the coordinates of the foot of the perpendicular from the point \((2,-6)\) to the line \(3 y-x+2=0\).

A curve is given by the parametric equations \(x=t^{2}-3, \quad y=t\left(t^{2}-3\right)\). (a) Find its Cartesian equation, in a form clear of surds and fractions. (b) Prove that it is symmetrical about the \(x\)-axis. (c) Show that there are no points on the curve for which \(x<-3\). (C)

Given the points \(\mathrm{A}(2,14), \mathrm{B}(-6,2), \mathrm{C}(12,-10)\) verify that the triangle \(\mathrm{ABC}\) is right angled. Calculate the coordinates of: (a) the point \(\mathrm{D}\) on \(\mathrm{AB}\) produced such that \(\mathrm{AC}=\mathrm{CD}\), (b) the point of intersection of the perpendiculars from \(A, C, D\) to the opposite sides of the triangle ACD.

In an attempt to find the condition that the line \(y=m x+c\) should touch the curve \(y^{2}+4 x=0\), the following solution was given. State the line where an error first appears. (a) The line meets the curve where \(y^{2}+4\left(\frac{y-c}{m}\right)=0\). (b) i.e. where \(m y^{2}+4 y-4 c=0\). (c) If the line is a tangent this equation has real roots. (d) The required condition is therefore \(m c=1\).
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