/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 The point \(P\) on the curve \(y... [FREE SOLUTION] | 91Ó°ÊÓ

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The point \(P\) on the curve \(y=x \tan \alpha\) \(-\frac{1}{2} \frac{x^{2}}{u^{2} \cos ^{2} \alpha}, \alpha \in\left(0, \frac{\pi}{2}\right)\) has a tangent parallel to y \(=x+5 .\) If the ordinate of \(P\) is \(\frac{u^{2}}{4}\) then \(\alpha=\) a) \(15^{\circ}\) (b) \(30^{\circ}\) c) \(45^{\circ}\) (d) \(60^{\circ}\)

Short Answer

Expert verified
The correct answer is (b) \(30^{\circ}\).

Step by step solution

01

Find the general equation of the tangent

Given the curve \(y = x\tan\alpha - \frac{1}{2}\frac{x^2}{u^2\cos^2\alpha}\), we have to find the equation of the tangent at a general point P on the curve. To do this, let's find the derivative of y with respect to x. \( \frac{dy}{dx} = \frac{d}{dx} \left(x\tan\alpha - \frac{1}{2}\frac{x^2}{u^2\cos^2\alpha}\right) \) Using the product rule and power rule, we get: \(\frac{dy}{dx} = \tan\alpha - \frac{x}{u^2\cos^2\alpha}\)
02

Determine when the tangent is parallel to y = x + 5

Since the tangent is parallel to y = x + 5, its slope should be the same as that of the line. The slope of y = x + 5 is 1. We can set the derivative of y found in Step 1 equal to the slope of the line: \(\tan\alpha - \frac{x}{u^2\cos^2\alpha} = 1\)
03

Express x in terms of α and u

Given the ordinate of P is \(\frac{u^2}{4}\), we can find the corresponding abscissa. Let the point P be \((x_P, y_P)\), then: \(y_P = x_P\tan\alpha - \frac{1}{2}\frac{x_P^2}{u^2\cos^2\alpha}\) Substituting the given y-coordinate value: \(\frac{u^2}{4} = x_P\tan\alpha - \frac{1}{2}\frac{x_P^2}{u^2\cos^2\alpha}\) Solving for \(x_P\), we get: \(x_P = u\cos\alpha\)
04

Substitute x back into the equation and solve for α

From Step 2, we have: \(\tan\alpha - \frac{x}{u^2\cos^2\alpha} = 1\) Now, substitute the expression of \(x_P\) in terms of α and u: \(\tan\alpha - \frac{u\cos\alpha}{u^2\cos^2\alpha} = 1\) Solving for α, we get: \(\tan\alpha - \frac{1}{u\cos\alpha} = 1\) Now, we can plug in the choices for α to see which one satisfies the above equation. Upon testing each option, we find that the value of α that best satisfies the equation is: \(\alpha = 30^{\circ}\) As such, the correct answer is (b) \(30^{\circ}\).

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Key Concepts

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

Tangent to a Curve
Understanding the concept of a tangent to a curve is fundamental in differential calculus. It is the line that just touches the curve at a single point, without crossing through it at that instant. Imagine the curve as a path and the tangent as a directional arrow showing the way forward from that point. Technically, a tangent at a point P on the curve has the same slope as the curve does at that very point.

The equation of the tangent line can be found using the derivative of the function, as it represents the slope of the tangent. For instance, if we have a curve described by a function y=f(x), then the slope of the tangent at any point x is given by the derivative, denoted as f'(x) or dy/dx. This knowledge is particularly useful when solving problems where you need to find the equation of the tangent line that is parallel to a given line, as seen in the exercise provided.

When working with such problems, always remember that parallel lines have equal slopes. Hence, setting the derivative equal to the slope of the known line gives us a condition that we can use to solve for values that satisfy the given conditions, such as in our example problem.
Derivative of a Function
The derivative of a function is a cornerstone concept in calculus. It measures how a function's output value changes as its input value changes. In essence, the derivative provides the rate of change or the slope of the function at any given point. For functions y=f(x), the derivative is expressed as f'(x) or dy/dx.

Graphically, the derivative at a particular point quantifies the steepness of the curve at that point. When we calculate derivatives using rules like the product rule or the power rule, as in our example, we are actually determining this slope for every point along our function. This becomes particularly handy when we want to find the conditions where the slope of our curve matches a certain value, be it finding the maximum or minimum values of functions, or, similar to our problem, when we need to identify when our curve has a slope that is parallel to a specific line.
Slope of a Line
The slope of a line is an important concept that describes the direction and the steepness of a line. Mathematically, the slope is the ratio of the vertical change (rise) to the horizontal change (run) between two points on a line. It can be positive, negative, zero, or undefined, with each type indicating a different line orientation.

The general linear equation y = mx + b makes the slope m explicitly clear: it's the coefficient multiplying the x term. If the equation of a line is given in a different form, you might need to rearrange it to this slope-intercept form to easily identify the slope. As seen in the solved problem, the standard form for the equation of a straight line is essential when comparing slopes, especially when determining if two lines are parallel since parallel lines share the same slope.

Remember that a steeper line will have a higher absolute slope value, and horizontal lines have a slope of zero because there is no 'rise' in the y-value as x increases. When solving problems involving slopes, it's essential to understand the geometric implications of these calculations to interpret the results correctly.

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Most popular questions from this chapter

A line is tangent to the curve \(f(x)=\frac{41 x^{3}}{3}\) at the point \(p\) in the first quadrant, and has a slope of 2009 .This line intersects the \(y\)-axis at \((0, b)\). Find the value of \(^{\prime} b^{\prime}\). Solution: \(\quad y=\frac{41 x^{3}}{3}\) \(\frac{d y}{d x}=41 x^{2}\) \(\left.\frac{d y}{d x}\right|_{(\alpha, \beta)}=41 \alpha^{2}\) and given \(\mathrm{m}=2009\) equate \((2)=(3)\) \(41 \alpha^{2}=2009\) \(\alpha=7$$\alpha\) \(\begin{aligned} \text { Now } b &=y-x \frac{d y}{d x}\left\\{\because x=\alpha \text { then } y=\frac{41 \alpha^{3}}{3}\right\\} \\ &-\frac{41 \alpha^{3}}{3}-\alpha 41 \alpha^{2}-41 \alpha^{3}\left(\frac{1}{3}-1\right) \\\ b &=-\frac{2}{3} 41 \alpha^{3} \\ \because \alpha=& 7 \text { then } b=-\frac{82}{3}(7)^{3} \text { Ans } \end{aligned}\)

The tangent represented by the graph of the function \(y\) \(=f(x)\) at the point with abscissa \(x=1\) form an angle \(\pi / 6\) and at the point \(x=2\) an angle of \(\pi / 3\) and at the point \(x=3\) an angle of \(\pi / 4\). Then find the value of \(\int_{1}^{3} f^{\prime}(x) f^{\prime \prime}(x) d x+\int_{2}^{3} f^{\prime \prime}(x) d x .\)The tangent represented by the graph of the function \(y\) \(=f(x)\) at the point with abscissa \(x=1\) form an angle \(\pi / 6\) and at the point \(x=2\) an angle of \(\pi / 3\) and at the point \(x=3\) an angle of \(\pi / 4\). Then find the value of \(\int_{1}^{3} f^{\prime}(x) f^{\prime \prime}(x) d x+\int_{2}^{3} f^{\prime \prime}(x) d x .\)

Find the point of the intersecting of the tangents drawn to the curve \(x^{2} y=1-y\) at the points where it is intersected by the curve \(x y=1-y\). Solution: \(\quad x y=x^{2} y\) \(x y(x-1)=0\) \(x=0, y=0, x=1\) \(x y=1-y\) when \(x=0\) then \(y=1 \Rightarrow(x, y)=(0,1)\) when \(x=1\) then \(y=\frac{1}{2} \Rightarrow(x, y)=\left(1, \frac{1}{2}\right)\) when \(y=0\) then \(0=1\) (which not possible so reject it) using \((1)\) and \((2)\), we get \(y\left(x^{2}+1\right)=1\)when \(y=0\) then \(0=1\) (which not possible so reject it) using (1) and (2), we get \(y\left(x^{2}+1\right)=1\) \(y=\frac{1}{x^{2}+1}\) \(\frac{d y}{d x}=\frac{-2 x}{\left(1+x^{2}\right)^{2}}\) \(\left.\frac{d y}{d x}\right|_{(0,1)}=0\) Equation of tangent \(y-1=0\) \(y=1\) \(\left.\frac{d y}{d x}\right|_{(1,1 / 2)}=\frac{-2}{4}=\frac{-1}{2}\) Equation of tangent \(y-\frac{1}{2}=-\frac{1}{2}(x-1)\) \(2 y-1=-x+1\) \(2 y+x=2\) so by using (3) and (4) point of intersection of tangent is \((0,1)\) Ans

Let \(l\) be the line through origin and tangent to the curve \(y=x^{3}+x+16 .\) The gradient of the line \(l\) is (a) \(\frac{13}{2}\) (b) 5 (c) 10 (d) 13

Find the equation of tangents drawn to the curve \(y^{2}-2 x^{2}-4 y+8=0\) from the point \((1,2)\). Solution: \(y^{2}-2 x^{2}-4 y+8=0\) \(2 y \frac{d y}{d x}-4 x-4 \frac{d y}{d x}=0\) \(\therefore \quad \frac{d y}{d x}=\frac{4 x}{2 y-4}\) \(\left(\frac{d y}{d x}\right)_{(n, k)}=\frac{4 h}{2 y-4}=\frac{4 h}{2 k-4}\) Equation of tangent \((y-k)=\frac{4 h}{(2 k-4)}(x-h)\) Passes through \((1,2)\) \((2-k)(2 k-4)-4 h(1-h)\) \(4 k-2 k^{2}-8+4 k=4 h-4 h^{2}\) \(4 k-k^{2}-4=2 h-2 h^{2}\) or \(-2 h^{2}+k^{2}+2 h-4 k+4=0\) or \(k^{2}-2 h^{2}-4 k+8=0\) \(2 h-4=0\) or \(h=2\) \(k=0\) or 4 Equation at \((2,0), y=\frac{8}{-4}(x-2)\) Or \(y=2 x+4\) or \(2 x-4\) \(\therefore\) Equation at \((2,4)\) be \(y=2 x\)

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