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Given that π₯ is equal to two π‘ cubed minus nine and π¦ is equal to the cube root of seven π‘ plus eight, determine the equation of the tangent to the curve at π‘ is equal to negative one.
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The question gives us a pair of parametric equations.
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And it wants us to determine the equation of the tangent to the curve at the point where π‘ is equal to negative one.
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We recall we can find the equation of a line with a slope of π which passes through the point π₯ one, π¦ one by using π¦ minus π¦ one is equal to π times π₯ minus π₯ one.
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So to find the equation of the tangent line to the curve, we need to find a point which it passes through and the slope of the tangent.
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We can find the point that our tangent line passes through by just substituting π‘ is equal to negative one into our pair of parametric equations.
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We also know the slope of the tangent is given by the change in π¦ with respect to π₯.
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And we recall, for a pair of parametric equations π₯ is equal to π of π‘ and π¦ is equal to π of π‘, we can find dπ¦ by dπ₯ by dividing the derivative of π¦ with respect to π‘ by the derivative of π₯ with respect to π‘.
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So we now know how to find the slope of our tangent and a point which our tangent passes through.
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Letβs start by finding the slope of our tangent.
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We need to find dπ¦ by dπ‘ and dπ₯ by dπ‘.
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We have dπ¦ by dπ‘ is equal to the derivative of the cube root of seven π‘ plus eight with respect to π‘.
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We see that this is the composition of two functions.
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So if we let π’ be equal to seven π‘ plus eight, we have π¦ is equal to the cube root of π’, and π’ is a function of π‘.
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So we can use the chain rule.
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The chain rule tells us the derivative of π¦ with respect to π‘ is equal to dπ¦ by dπ’ multiplied by dπ’ by dπ‘, if π¦ is a function of π’ and π’ is a function of π‘.
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We see the cube root of π’ is just equal to π’ to the power of one-third.
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So we can differentiate this by using the power rule for differentiation.
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We multiply by the exponent and then reduce the exponent by one.
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We now need to multiply this by the derivative of π’ with respect to π‘.
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We can differentiate seven π‘ plus eight with respect to π‘ by using the power rule for differentiation.
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Itβs just equal to seven.
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Finally, we can rewrite dπ¦ by dπ‘ in terms of π‘ by substituting π’ is equal to seven π‘ plus eight.
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So by using this substitution and rearranging our expression, we have dπ¦ by dπ‘ is equal to seven divided by three times seven π‘ plus eight to the power of two over three.
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We can do the same to find dπ₯ by dπ‘.
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Itβs equal to the derivative of two π‘ cubed minus nine with respect to π‘.
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We can evaluate this by using the power rule for differentiation.
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Itβs equal to six π‘ squared.
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We can use these to find an expression for the slope dπ¦ by dπ₯.
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Itβs equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘ or dπ¦ by dπ‘ multiplied by the reciprocal of dπ₯ by dπ‘.
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This gives us dπ¦ by dπ₯ is equal to seven divided by three times seven π‘ plus eight to the power of two over three multiplied by the reciprocal of six π‘ squared.
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We can then use this to find the slope of our tangent line when π‘ is equal to negative one.
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We substitute π‘ is equal to negative one into this expression for the slope.
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This gives us a slope of seven divided by three times seven times negative one plus eight to the power of two over three multiplied by the reciprocal of six times negative one squared.
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We have seven times negative one plus eight is just equal to one.
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And then one to the power of two-thirds is just equal to one.
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And then negative one squared is just equal to one.
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And so the reciprocal of six times one is just one-sixth.
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So this simplifies to give us the slope of our tangent is seven-thirds times one-sixth which we can calculate to give us seven divided by 18.
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So we found the slope of our tangent line.
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We now need to find the π₯- and π¦-coordinate of a point our line passes through.
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To do this, weβll substitute π‘ is equal to negative one into each of our parametric equations.
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Substituting π‘ is equal to negative one into our parametric equation for π₯ gives us π₯ is equal to two times negative one cubed minus nine, which we can calculate to be negative 11.
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Substituting π‘ is equal to negative one into our parametric equation for π¦ gives us π¦ is equal to the cube root of seven times negative one plus eight, which is, of course, just equal to one.
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So our tangent line passes through the point negative 11, one and has a slope of seven divided by 18.
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Using our formula for the equation of our line, we have the equation of our tangent when π‘ is equal to negative one is given by π¦ minus one is equal to seven over 18 multiplied by π₯ minus negative 11.
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We can simplify this equation by multiplying both sides through by 18.
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Multiplying π¦ minus one by 18 gives us 18π¦ minus 18.
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And since seven over 18 multiplied by 18 is just equal to seven, we have the right-hand side of our equation is seven multiplied by π₯ minus negative 11.
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We know that subtracting negative 11 is the same as adding 11.
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This gives us 18π¦ minus 18 is equal to seven times π₯ plus 11.
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Multiplying out the parentheses, rearranging, and simplifying, we can show that the equation of our line is given by negative seven π₯ plus 18π¦ minus 95 is equal to zero.
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Therefore, weβve shown the equation to the tangent of the curve π₯ is equal to two π‘ cubed minus nine and π¦ is equal to the cube root of seven π‘ plus eight when π‘ is equal to negative one.
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Is given by the equation negative seven π₯ plus 18π¦ minus 95 is equal to zero.