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Let xx and yy be functions of tt with y=x13y = x^{\frac{1}{3}}. If dxdt=7\frac{dx}{dt} = 7, what is dydt\frac{dy}{dt} when x=8x = 8?\newlineWrite an exact, simplified answer.

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Q. Let xx and yy be functions of tt with y=x13y = x^{\frac{1}{3}}. If dxdt=7\frac{dx}{dt} = 7, what is dydt\frac{dy}{dt} when x=8x = 8?\newlineWrite an exact, simplified answer.
  1. Identify Relationship: Identify the relationship and differentiate using the chain rule.\newlineGiven y=x13y = x^{\frac{1}{3}}, differentiate both sides with respect to tt.\newlinedydt=13x23dxdt\frac{dy}{dt} = \frac{1}{3}x^{-\frac{2}{3}} \cdot \frac{dx}{dt}
  2. Differentiate Using Chain Rule: Substitute the given values into the differentiated equation.\newlinedxdt=7\frac{dx}{dt} = 7 and x=8x = 8.\newlinedydt=(13)(8)23×7\frac{dy}{dt} = \left(\frac{1}{3}\right)(8)^{-\frac{2}{3}} \times 7
  3. Substitute Given Values: Calculate the value of x(2/3)x^{(-2/3)} when x=8x = 8.\newline8(2/3)=1/(8(2/3))=1/(4)=0.258^{(-2/3)} = 1/(8^{(2/3)}) = 1/(4) = 0.25
  4. Calculate x23x^{-\frac{2}{3}}: Finish the calculation for dydt\frac{dy}{dt}.
    dydt=(13)0.257=0.583333\frac{dy}{dt} = \left(\frac{1}{3}\right) * 0.25 * 7 = 0.583333\ldots

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