Question Video: Finding the Equation of a Curve given the Expression of the Slope of Its Tangent and a Point on the Curve Mathematics

Find the equation of the curve that passes through the point (βˆ’2, 1) given that the gradient of the tangent to curve is βˆ’11π‘₯Β².

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Video Transcript

Find the equation of the curve that passes through the point negative two, one given that the gradient of the tangent to the curve is negative 11π‘₯ squared.

In this question, we want to find the equation of a curve. And to do this, we’re given some information about our curve. We’re told that our curve passes through the point negative two, one and the gradient of the tangent to this curve is equal to the function negative 11π‘₯ squared. So let’s break down the two pieces of information we’re given about our curve. First, if we wanted to write the equation of this curve in the form 𝑦 is equal to 𝑓 of π‘₯, then saying that our curve passes through the point negative two, one means when π‘₯ is equal to negative two, 𝑦 should be equal to one.

So we can substitute these values into the equation for our curve. We should have that one is equal to 𝑓 evaluated at negative two. Next, we’re told the gradient of the tangent to our curve is negative 11π‘₯ squared. And remember, we know the gradient of the tangent to our curve at a value of π‘₯ is as rate of change of 𝑦 with respect to π‘₯. In other words, this is telling us d𝑦 by dπ‘₯ is equal to negative 11π‘₯ squared. So we need to find the equation of a curve given its slope and a point on the curve. And to do this, we’re going to need to find an antiderivative of negative 11π‘₯ squared. And we know how to find antiderivatives by using integration. The integral of negative 11π‘₯ squared with respect to π‘₯ will give us the most general antiderivative of negative 11π‘₯ squared.

And remember, the reason we’re finding this is we know the derivative of our expression for 𝑦 with respect to π‘₯ has to be equal to negative 11π‘₯ squared. So we need to integrate negative 11π‘₯ squared with respect to π‘₯. We can do this by using the power rule for integration. We recall this tells us for any real constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Ž times π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐢. We add one to our exponent and then divide by this new exponent.

In our case, the value of π‘Ž is negative 11 and the value of 𝑛 is two. So we add one onto our exponent of two to get an exponent of three and then divide by this new exponent. We get negative 11π‘₯ cubed over three plus a constant of integration 𝐢. And of course, the derivative of this expression with respect to π‘₯ is equal to negative 11π‘₯ squared. So far, for our curve 𝑦 is equal to negative 11π‘₯ cubed over three plus a constant of integration 𝐢, we have d𝑦 by dπ‘₯ is equal to negative 11π‘₯ squared. However, remember, we still need this curve to pass through the point negative two, one. And to do this, we need to substitute π‘₯ is equal to negative two and 𝑦 is equal to one in to our curve.

Substituting π‘₯ is equal to negative two and 𝑦 is equal to one into the equation for our curve, we get one is equal to negative 11 times negative two cubed all over three plus our constant of integration 𝐢. Now all we need to do is solve this equation for 𝐢. Rearranging this equation, we see that 𝐢 should be equal to negative 85 divided by three. Now all we need to do is use this value of 𝐢 in the equation for our curve, and this gives us our final answer. Therefore, we were able to show the equation of the curve that passes through the point negative two, one and which has the gradient of its tangent line at π‘₯ given by negative 11π‘₯ squared has the equation 𝑦 is equal to negative 11 over three times π‘₯ cubed minus 85 over three.

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