Explanation Given information : Given parabola y = 1 − x 2 and the triangle A B C formed by the x − axis and the tangent line at P and Q is an equilateral triangle. From the given figure we can observe that the y = 1 − x 2 is symmetric about y − axis Which implies the tangents formed at the points P and Q on the parabola also symmetric about y − axis . Which implies, if Q = ( a , b ) be a point on the parabola then reflection of Q = ( a , b ) along y − axis is given by P = ( − a , b ) . Now we find equation of tangent passing through the point Q = ( a , b ) , Which is given by y − b = m ( x − a ) , where slope of the tangent, which is equal to derivative of parabola at Q = ( a , b ) ⇒ y | = − 2 x ⇒ m = − 2 a Thus, the required tangent equation A C is y − b = − 2 a ( x − a ) Where the coordinate of point C is when the tangent intersect at x − axis . ⇒ − b = 2 a ( x − a ) as y = 0 on x − axis ⇒ − b = 2 a x − 2 a 2 x = 2 a 2 − b 2 a Thus, the coordinate of point C is ( 2 a 2 − b 2 a , 0 ) . Similar way the coordinate of point A is when the tangent intersect at y − axis . ⇒ y − b = 2 a ( 0 − a ) as x = 0 on y − axis . ⇒ y − b = − 2 a 2 y = − 2 a 2 + b

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ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Chapter 3, Problem 93E

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The points P and Q on the parabola y=1−x2 so that the triangle ABC formed by the x−axis and the tangent line at P and Q is an equilateral triangle.

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Chapter 3, Problem 92E

Chapter 3, Problem 94E

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The correct answer is D Could you explain and show the steps please

Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

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Chapter 3 Solutions

Biocalculus

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The points P and Q on the parabola y = 1 − x 2 so that the triangle A B C formed by the x − axis and the tangent line at P and Q is an equilateral triangle.

Solution Summary: The author explains that the triangle ABC formed by the x-axis and the tangent line at P and Q is an equilateral triangle.

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The points P and Q on the parabola y=1−x2 so that the triangle ABC formed by the x−axis and the tangent line at P and Q is an equilateral triangle.

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