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Chapter 2: Problem 7

Determine the slope of the tangent to each of the curves at the given point. a. \(y=3 x^{4},(1,3)\) b. \(y=\frac{1}{x^{-5}},(-1,-1)\) c. \(y=\frac{2}{x},(-2,-1)\) d. \(y=\sqrt{16 x^{3}},(4,32)\)

Short Answer

Expert verified
The slopes are 12, 5, -1/2, and 12 for curves a, b, c, and d respectively.

Step by step solution

01

Find the derivative for equation a

The function is given by \( y = 3x^4 \). To find the derivative, use the power rule which states that the derivative of \( x^n \) is \( nx^{n-1} \). Therefore, \( y' = 4 \cdot 3x^{4-1} = 12x^3 \). This function represents the slope of the tangent at any point \( x \).
02

Evaluate the slope for equation a at given point

Substitute \( x = 1 \) into the derivative \( y' = 12x^3 \) to find the slope of the tangent at point \((1, 3)\). This gives \( y' = 12(1)^3 = 12 \). Thus, the slope at point \((1,3)\) is \( 12 \).
03

Find the derivative for equation b

The function is given by \( y = \frac{1}{x^{-5}} \). Simplifying, \( y = x^{5} \). Using the power rule, the derivative is \( y' = 5x^{4} \).
04

Evaluate the slope for equation b at given point

Substitute \( x = -1 \) into \( y' = 5x^{4} \) to find the slope of the tangent at point \((-1, -1)\). This gives \( y' = 5(-1)^4 = 5 \). Thus, the slope at point \((-1,-1)\) is \( 5 \).
05

Find the derivative for equation c

The function is \( y = \frac{2}{x} \), which can be rewritten as \( y = 2x^{-1} \). Using the power rule, the derivative becomes \( y' = -2x^{-2} \) or \( y' = -\frac{2}{x^2} \).
06

Evaluate the slope for equation c at given point

Substitute \( x = -2 \) into \( y' = -\frac{2}{x^2} \) to find the slope of the tangent at point \((-2, -1)\). This calculates to \( y' = -\frac{2}{(-2)^2} = -\frac{2}{4} = -\frac{1}{2} \). Thus, the slope at point \((-2,-1)\) is \( -\frac{1}{2} \).
07

Find the derivative for equation d

The function is \( y = \sqrt{16x^3} = (16x^3)^{1/2} \). Use the chain rule and power rule to differentiate: \( y' = \frac{1}{2}(16x^3)^{-1/2} \cdot 48x^2 = \frac{24x^2}{\sqrt{16x^3}} \). Simplifying, this becomes \( y' = \frac{24x^2}{4x^{3/2}} = 6x^{1/2} \).
08

Evaluate the slope for equation d at given point

Substitute \( x = 4 \) into \( y' = 6x^{1/2} \) to find the slope at point \((4, 32)\). This yields \( y' = 6 \times 4^{1/2} = 6 \times 2 = 12 \). Thus, the slope at point \((4,32)\) is \( 12 \).

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Key Concepts

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

Tangent Line Slope
Finding the slope of a tangent line to a curve at a given point is a crucial concept in calculus. The slope tells us how steep the line is at that specific point.
  • The tangent line "touches" the curve at precisely one point and has the same slope as the curve at that location.
  • To find the slope of a tangent line, we first need the derivative of the function, as it represents the slope of the curve at any point.
  • Once we have the derivative, we can substitute the x-coordinate of the given point into the derivative to find the slope at that specific point.
Understanding the concept of tangent line slope is essential as it links the dynamic change of a function to its static representation on a graph.
Power Rule
The Power Rule is a quick shortcut for differentiating functions of the form \( x^n \). It is one of the first derivative rules taught in calculus and is widely used because of its simplicity.To apply the power rule, follow these steps:
  • Identify the exponent \( n \) in the expression \( x^n \).
  • Multiply the entire expression by the exponent, resulting in \( n \cdot x^{n-1} \).
  • Subtract one from the original exponent to finish the derivation: \( n-1 \).
Using the power rule, derivatives can be determined rapidly, which is crucial for efficiently finding tangent lines and evaluating limits in other calculus problems.
Chain Rule
The Chain Rule is an essential technique for finding the derivatives of composite functions. A composite function is a function nested inside another function, such as \( (g(x))^n \). The chain rule allows us to take these apart for differentiation.When applying the chain rule:
  • Differentiating the outer function first and leave the inner function untouched initially.
  • Next, take the derivative of the inner function.
  • Multiply the derivative of the outer function by the derivative of the inner function to get the composite derivative.
An example from the exercise is \( \sqrt{16x^3} \). It uses both the power rule and the chain rule to manage the nested structure effectively.
Derivative Evaluation
Evaluating derivatives at a specific point allows us to find the slope of the tangent line or the rate of change of the function at that point. It involves substituting the x-coordinate value into the derivative function. Consider these steps to perform a derivative evaluation:
  • First, ensure you have the correct derivative of the function.
  • Substitute the x-value of the point you're interested in.
  • Compute the result to find the slope of the tangent line at the point.
The final number represents how much the function is changing at that exact point 鈥� an indispensable tool in calculus for analyzing functions and their behaviors.

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Most popular questions from this chapter

Illustrate two situations in which a function does not have a derivative at \(x=1\)

For what values of \(x\) do the curves \(y=\left(1+x^{3}\right)^{2}\) and \(y=2 x^{6}\) have the same slope?

Show that there are two tangents to the curve \(y=\frac{1}{5} x^{5}-10 x\) that have a slope of 6.

For each of the following pairs of functions, find the composite functions \((f \circ g)\) and \((g \circ f) .\) What is the domain of each composite function? Are the composite functions equal? a. \(f(x)=x^{2}\) \(g(x)=\sqrt{x}\) b. \(f(x)=\frac{1}{x}\) \(g(x)=x^{2}+1\) c. \(f(x)=\frac{1}{x}\) \(g(x)=\sqrt{x+2}\)

a. The radius of a circular juice blot on a piece of paper towel \(t\) seconds after it was first seen is modelled by \(r(t)=\frac{1+2 t}{1+t},\) where \(r\) is measured in centimetres. Calculate i. the radius of the blot when it was first observed ii. the time at which the radius of the blot was \(1.5 \mathrm{cm}\) iii. the rate of increase of the area of the blot when the radius was \(1.5 \mathrm{cm}\) b. According to this model, will the radius of the blot ever reach \(2 \mathrm{cm} ?\) Explain your answer.
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