Video Transcript
The gradient of the tangent of a curve is negative four π₯ plus four all divided by three π¦ plus three and the curve passes through the point negative two, negative three. Find the equation of the normal to the curve at the point where the π₯-coordinate is negative two.
In this question, weβre given the gradient of the tangent of the curve in terms of π₯ and π¦. Remember, that will tell us the slope of the curve. And weβre also told that the curve passes through the point negative two, negative three. We need to determine the equation of the normal to this curve at the point where the π₯-coordinate is negative two. And there appears to be a very simple way to answer this question.
We have a point on the curve with π₯-coordinate negative two. We want to find the equation of the normal to the curve when π₯ is equal to negative two. And we can find the slope of the tangent at this point by using the given expression. Then, the slope of the normal will be negative the reciprocal of this value, and this would work. We could use this to find the equation of the normal to this curve at the point negative two, negative three.
However, thereβs one assumption weβre making when we use this method. Weβre assuming that the only point on our curve with π₯-coordinate negative two is the point negative two, negative three. Another way of saying this is weβre assuming weβre graphing a function. However, we can show this is not the case. To do this, letβs start with the given equation for the slope of this curve. dπ¦ by dπ₯ is equal to negative four π₯ plus four all divided by three π¦ plus three. We want to use this to find an equation for π¦. And since weβre given the slope of this curve, thatβs the rate of change of π¦ with respect to π₯, we need to find an antiderivative. Weβre going to do this by using indefinite integrals.
So letβs rearrange this equation. Letβs start by multiplying through by three π¦ plus three. Doing this and then distributing over our parentheses, we get three π¦ dπ¦ by dπ₯ plus three dπ¦ by dπ₯ is equal to negative four π₯ plus four. Weβre now going to integrate both sides of this equation with respect to π₯. And remember, we can integrate each term separately. This gives us the integral of three π¦ dπ¦ by dπ₯ with respect to π₯ plus the integral of three dπ¦ by dπ₯ with respect to π₯ is equal to the integral of negative four π₯ plus four with respect to π₯.
We now need to evaluate each of these integrals separately. In our first integral, we can see the expression π¦ multiplied by dπ¦ by dπ₯. And we recall finding the indefinite integral of an expression means weβre finding an antiderivative of that expression. And we can also recall the following application of the chain rule. The derivative of π¦ squared with respect to π₯ will be equal to two multiplied by π¦ multiplied by dπ¦ by dπ₯. In other words, we can use this to find an antiderivative of our expression.
The only difference between these two expressions is the constant coefficient at the start of the expression. So we can just multiply both sides of our equation by three over two. Therefore, what weβve shown is if we differentiate three over two π¦ squared with respect to π₯, we get three π¦ dπ¦ by dπ₯. In other words, this is an antiderivative of this expression. And remember, we do get a constant of integration when we evaluate this integral. However, we can combine all of the constants of integration at the end of our expression.
Now, letβs move on to evaluating our second integral. Since the indefinite integral of an expression gives us the most general antiderivative of this expression, we can see this is three π¦, since the derivative of three π¦ with respect to π₯ is three times dπ¦ by dπ₯. We can now move on to our third indefinite integral. Since this is the integral of a polynomial expression in π₯, we can use the power rule for integration. We recall this tells us for any real constants π and π, where π is not equal to negative one, the integral of ππ₯ to the πth power with respect to π₯ is equal to π multiplied by π₯ to the power of π plus one divided by π plus one plus the constant of integration πΆ. We add one to the exponent and divide by this new exponent.
We can apply this to integrate the first term. We write π₯ as π₯ to the first power, add one to the exponent to get a new exponent of two, and divide by this new exponent. This gives us negative four π₯ squared over two. We could do the same to evaluate the integral of our second term, since four is four times π₯ to the zeroth power. However, itβs easier to remember the slope of a linear function is given by the constant coefficient of the variable. In particular, the derivative of four π₯ will be four. So four π₯ is an antiderivative of four. Finally, remember, we need to add a constant of integration πΆ. Simplifying this, weβve now shown that three over two π¦ squared plus three π¦ is equal to negative two π₯ squared plus four π₯ plus πΆ.
And we can actually simplify this equation slightly. Since all of our coefficients are integers except for the first one, which is rational, we can just multiply both sides of this equation through by two. This gives us three π¦ squared plus six π¦ is equal to negative four π₯ squared plus eight π₯ plus two πΆ. However, remember that πΆ is a constant of integration, so two πΆ is also a constant. So we can relabel this. For simplicity, weβll relabel this πΆ. Letβs clear some space now that we found an equation for our curve. Remember, we found this equation to try and determine all of the points on the curve with π₯-coordinate negative two. This means we would need to substitute π₯ is equal to negative two into this equation and solve for π¦.
However, thereβs a problem. We still have an unknown value of πΆ. But we can find this value of πΆ since weβre told in the question the point negative two, negative three lies on the curve. Therefore, we can substitute π¦ is equal to negative three and π₯ is equal to negative two into this equation to find the value of πΆ. This gives us three times negative three squared plus six times negative three is equal to negative four multiplied by negative two squared plus eight multiplied by negative two plus πΆ.
Evaluating each term gives us 27 minus 18 is equal to negative 16 minus 16 plus πΆ. And if we evaluate and rearrange for πΆ, we see that πΆ will be equal to 41. We can then substitute this value of πΆ back into the equation of our curve. This then gives us an equation for our curve entirely in terms of the variables π₯ and π¦.
We can now determine every point on this curve with π₯-coordinate negative two by substituting π₯ is equal to negative two into the curve and solving for π¦. Substituting π₯ is equal to negative two into the equation of the curve, we get three π¦ squared plus six π¦ is equal to negative four multiplied by negative two squared plus eight times negative two plus 41. Evaluating this and simplifying, we get three π¦ squared plus six π¦ is equal to nine, which is a quadratic equation in π¦. We can simplify this slightly by noticing all three terms share a factor of three. Rearranging this gives us the equation π¦ squared plus two π¦ minus three is equal to zero. We can solve this by factoring.
Since three multiplied by negative one is negative three and three plus negative one is equal to two, we can factor this to get π¦ plus three multiplied by π¦ minus one is equal to zero. And for the product of two factors to be equal to zero, one of the two factors must be equal to zero. Therefore, we get two possible π¦-coordinates of points on our curve with π₯-coordinate negative two. Either π¦ is negative three or π¦ is one. Therefore, we need to find the equations of the normals to our curve at both of these two points.
To do this, letβs start by doing this with the point given to us in the question, the point negative two, negative three. First, letβs substitute the coordinates of this point into our expression to find the slope of the tangent line at this point. This gives us the slope of the tangent line at this point is equal to negative four times negative two plus four all divided by three multiplied by negative three plus three.
And we can evaluate this expression. If we do, weβll see that itβs equal to negative two. But remember, this is the gradient of the tangent line at this point. We need to find the gradient of the normal line. And we can find this by recalling the slope of the normal line will be negative the reciprocal of the slope of the tangent line at this point. In other words, this is negative one times negative two raised to the power of negative one, which we can calculate is equal to one-half.
Therefore, we found the slope of this normal line. And we also know that this normal line passes through the point negative two, negative three. This means we can find the equation of this normal line. Weβll use the pointβslope form of the equation of a line. Thatβs the form π¦ minus π¦ one is equal to π times π₯ minus π₯ one, where the line passes through the point with coordinates π₯ one, π¦ one and has a slope of π.
Substituting in the coordinates of the point and the slope of one-half, we get π¦ minus negative three is equal to one-half times π₯ minus negative two. And we can then simplify and rearrange this equation into the general equation of a straight line. We get two π¦ minus π₯ plus four is equal to zero. But this only gives us the equation of our first normal line. We also need to find the equation of the normal line at the point negative two, one.
So letβs clear some space and follow the same process to find an equation for the normal line at this point. Weβll start by substituting π₯ is negative two and π¦ is one into our expression for the slope of the tangent line. If we do this, we get negative four times negative two plus four all divided by three times one plus three, which we can evaluate. We see that itβs equal to two. The slope of the normal line at this point will be negative the reciprocal of this value. And negative the reciprocal of two is negative one-half. Finally, we can substitute this value for the slope and the coordinates of this point into the equation of our line to find the equation of the normal line at this point.
We get π¦ minus one is equal to negative one-half multiplied by π₯ minus negative two. And if we simplify this equation and rewrite it in the general form, we get two π¦ plus π₯ is equal to zero. This then gives us the equations of both normal lines for the two points on our curve with π₯-coordinate negative two. These have equations two π¦ plus π₯ is equal to zero and two π¦ minus π₯ plus four is equal to zero.