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It is given that \( y=\frac{\ln \left(3 x^{2}-1\right)}{x+2} \), for \( x>\frac{1}{\sqrt{3}} \). When \( x=1, y \) is increasing at the rate of \( h \) units per second. Find, in terms of \( h \), the corresponding rate of change in \( x \), giving your answer in exact form.

Ask by Mckenzie Turnbull. in Mauritius
Mar 15,2025

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The rate of change in \( x \) is \( \frac{9h}{9-\ln2} \) units per second.

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To find the rate of change of \( x \) in terms of \( h \) when \( x = 1 \), we start by differentiating \( y \) with respect to \( x \). Given \( y = \frac{\ln(3x^2 - 1)}{x + 2} \), we apply the quotient rule for differentiation: \[ \frac{dy}{dx} = \frac{(x + 2) \cdot \frac{d}{dx}[\ln(3x^2 - 1)] - \ln(3x^2 - 1) \cdot \frac{d}{dx}[x + 2]}{(x + 2)^2} \] Calculating \( \frac{d}{dx}[\ln(3x^2 - 1)] \) using the chain rule gives: \[ \frac{d}{dx}[\ln(3x^2 - 1)] = \frac{1}{3x^2 - 1} \cdot (6x) = \frac{6x}{3x^2 - 1} \] The differentiated form of \( x + 2 \) is simply \( 1 \). Hence, substituting in the quotient rule: \[ \frac{dy}{dx} = \frac{(x + 2) \cdot \frac{6x}{3x^2 - 1} - \ln(3x^2 - 1) \cdot 1}{(x + 2)^2} \] \[ = \frac{(6x(x + 2)) / (3x^2 - 1) - \ln(3x^2 - 1)}{(x + 2)^2} \] Now, we find \( \frac{dy}{dx} \) at \( x = 1 \). Calculating \( y \) at \( x = 1 \): \[ y(1) = \frac{\ln(3(1)^2 - 1)}{1 + 2} = \frac{\ln(2)}{3} \] Then we need to compute the components of \( \frac{dy}{dx} \): 1. Evaluating \( 3x^2 - 1 \) at \( x = 1 \): \[ 3(1)^2 - 1 = 2 \] 2. Thus, \( \ln(3(1)^2 - 1) = \ln(2) \), and: \[ \frac{dy}{dx} = \frac{(6 \cdot 1 (1 + 2))/(2) - \ln(2)}{(1 + 2)^2} \] \[ = \frac{(18/2) - \ln(2)}{9} = \frac{9 - \ln(2)}{9} = 1 - \frac{\ln(2)}{9} \] Now we have that \( \frac{dy}{dx} = 1 - \frac{\ln(2)}{9} \) at \( x = 1 \). Using the chain rule to relate \( \frac{dy}{dt} \), \( \frac{dx}{dt} \), and \( \frac{dy}{dx} \): \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \] Since \( \frac{dy}{dt} = h \): \[ h = \left(1 - \frac{\ln(2)}{9}\right) \cdot \frac{dx}{dt} \] Rearranging gives: \[ \frac{dx}{dt} = \frac{h}{1 - \frac{\ln(2)}{9}} \] Thus, the corresponding rate of change in \( x \) in terms of \( h \) is: \[ \frac{dx}{dt} = \frac{9h}{9 - \ln(2)} \] This is the rate of change of \( x \) when \( x = 1 \), expressed in terms of \( h \).

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If you invest \( \$ 1000 \) for \( t \) years at \( 6.71 \% \) compounded at the frequencies below, find the following. (a) Suppose you compound at \( 6.71 \% \) monthly. i) Report an expression equivalent to the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded monthly by completing the box with the growth factor if compounded annually. 1000 \( \square \) Number \( t \) (Round to \( \underline{4} \) decimal places.) ii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (b) Suppose you compound at \( 6.71 \% \) continuously. i) You would expect \( 6.71 \% \) compounded continuously to give a \( \square \) Click for List yield than what is given in part (a). ii) Complete the boxes below to report the expression for the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded continuously and the equivalent growth factor if compounded annually. \[ \begin{array}{l} 1000 e^{(\text {Number } t)} \\ \approx 1000(\text { Number })^{t} \end{array} \] (Round to \( \underline{4} \) decimal places.) iii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (c) Complete the boxes to summarize: i) From part (a) we have that 6.71 \% compounded monthly is equivalent to \( \square \) Number \( \% \) compounded annually. ii) From part (b) we have that 6.71 \% compounded continuously is equivalent to \( \square \) Number \( \% \) compounded annually.

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If you invest \( \$ 1000 \) for \( t \) years at \( 6.71 \% \) compounded at the frequencies below, find the following. (a) Suppose you compound at \( 6.71 \% \) monthly. i) Report an expression equivalent to the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded monthly by completing the box with the growth factor if compounded annually. 1000 \( \square \) Number \( t \) (Round to \( \underline{4} \) decimal places.) ii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (b) Suppose you compound at \( 6.71 \% \) continuously. i) You would expect \( 6.71 \% \) compounded continuously to give a \( \square \) Click for List yield than what is given in part (a). ii) Complete the boxes below to report the expression for the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded continuously and the equivalent growth factor if compounded annually. \[ \begin{array}{l} 1000 e^{(\text {Number } t)} \\ \approx 1000(\text { Number })^{t} \end{array} \] (Round to \( \underline{4} \) decimal places.) iii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (c) Complete the boxes to summarize: i) From part (a) we have that 6.71 \% compounded monthly is equivalent to \( \square \) Number \( \% \) compounded annually. ii) From part (b) we have that 6.71 \% compounded continuously is equivalent to \( \square \) Number \( \% \) compounded annually.
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